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GROUPS OF ORDER °° 


BY 


MYRON OWEN TRIPP 


SUBMITTED IN PARTIAL FULFILMENT OF THE REQUIREMENTS FOR THE 
DeEGREE oF Doctor oF PHILOSOPHY, IN THE FACULTY OF 


Pure Science, CoLUMBIA UNIVERSITY 


PRESS OF 
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LANCASTER, PA 


1909 


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GROUPS OF ORDER p°¢ 
BY 


MYRON OWEN TRIPP 


ih 
INTRODUCTION. 


1.* Historical note. Cayley + called attention to an important desideratum 
in the theory of groups, viz., the determination of all groups of a given order 7; 
for n = 2, 3, 4, 5, 6 he found all the types of G,. A. B. Kempet enumer- 
ated the types of G.(n=1, 2, ---, 12) and gave a graphical representation in 
each case. CAYLEY § remarked that in studying types of groups up to order 
11 the first case that involves difficulty is G,;. He also called attention to 
the fact that Kempe (1. c.) made an error in enumerating the types of G,,. 
CaYLeEy and Kempe proceeded according to order, e. g., they treated G,, but 
did not deal with G,, in general. BuRNsIDE|| gives the number of distinct 
types for all orders less than 32. The determination of the number of types of 
G,, caused considerable discussion. Concerning these types LEVAVASSEUR J 
said, “‘I have already found more than 75 distinct groups, and I have not yet 
finished the enumeration.” Shortly afterwards MILLER announced ** that the 


. - number of these groups is 51. About two years later BaGNERA {+ stated that 


the number of G,, is only 50. Since then, however, he has conceded that 
Miller was correct in saying the number of these groups is 51. 

All G, are cyclic. The types of G,, and G,, aregiven by NEtto.{{ G,,,, 
G 24 and Gs have been discussed by CoLE and GLOVER, §§ while Gs and Gia have 
been treated by Younga.|||| A very important memoir is that of HOLDER YF on 





* Throughout this paper the letters p, g, r, ---, denote different prime numbers. A group 
of order p+g® --- is denoted by Ga, *--, while subgroups are denoted by H’s with subscripts 
to indicate their orders. 

tAmerican Journal of Mathematics (1878), vol. 1, pp. 50-52. 

t Philosophical Transactions (1886), vol. 177, pp. 1-70. 

§ American Journal of Mathematics (1888), vol. 2, pp. 139-157. 

|| Theory of Groups, p. 105 and chap. V. 

{Comptes Rendus (1896), vol. 122, p. 182. 

** Comptes Rendus (1896), vol. 122, p. 370. 

ttAnnali di Matematica (1898), p. 139. 

tt Substitution Groups, Cole’s Translation, p. 149. 

§§ American Journal of Mathematios (1°93), vol. 15, pp. 191-220. 

||| American Journal of Mathematics (1893), vol. 15, pp. 124-178. 

{] Mathematische Annalen (1893), vol. 43, pp. 301-412. 


3 


4 M. 0. TRIPP: GROUPS OF ORDER p*q’ 


groups of orders p*, pq’, pg’, p*. The others who have dealt with groups, whose 
orders are represented by four primes, are WESTERN * on G,, , LEVAVASSEUR { 
on Gi.2, HOLDER] on G,,,,..., GLENN§ on G,,,, and MILLER||{ on Ee 
and G5. 

The following have enumerated the types of groups whose orders are repre- 
sented by five primes: BAGNERA** on G,; LEVAVASSEUR}{ on G,,, (p odd). 

Recently Porron has given a list of the types of G,, in his Paris thesis. 

As regards the solubility of G2, it may be noted that this was proven by 
Sytow {{ for 8 = 0, by Fropentus§§ for 8 =1, by JorpDAn|||| for B= 2, 
by CoLEg{{ for 8 = 3, by BuRNsipE*** for all values of 8. 

Objects and results of the present investigation. The principal aim of this 
discussion is the determination of the defining relations for all distinct types 
of abstract G,,., no one of which is simply isomorphic with any other. As 
the number of primes, either the same or different, increases the problem 
complicates with remarkable rapidity. This is seen on comparing Hélder’s 
treatment of G,, with that of G,,. One of the most important parts of the 
process of obtaining types is the determination of the invariant subgroups 
necessary for defining the types, which frequently involves considerable diffi- 
culty. When one of the primes p or q is 2 the determination of the defining 
relations becomes more difficult, in general, than for larger values. This arises 
from the fact that the invariant subgroups which exist for a prime greater than 
2 do not necessarily exist when p equals 2. 

The number of H,, and H, is given for every type of G,,,.. Especial atten- 
tion is also given to decomposable groups, that is, those G,.. which can be 
formed by taking the direct product of two or more subgroups of lower order. 
Thus the defining relations of the decomposable groups may be checked by 
comparing with results previously worked out. The non-decomposable groups 
are checked by using all possible relations to discover if any inconsistency 
arises. In many cases, dependent on certain relations between p and q, the 
number of different types increases indefinitely as p or q increases. ‘This is 
not the case with groups of orders, p, p?, pq, p* or p*; but it is the case with 


* Proceedings of the London Mathematical Society (1899), vol. 30, pp. 209-263. 
tAnnales Scientifiques de 1’Ecole Normale Supérieure (1902), pp. 335-353. 
{Géttinger Nachrichten (1895), pp. 211-229. 7 
§ Transactions of the American Mathematical Society (1906), pp. 137-151. 

|| Philosophical Magazine (1896), vol. 42, pp. 195-200. 

{Quarterly Journal of Mathematics (1898), pp. 259-263. 

** Annali di Matematica (1898), pp. 137-228. 

tt Annales Toulouse (1903), pp. 63-123. 

tt Mathematische Annalen, vol. 5, p. 588. 

§§ Berliner Sitzungsberichte (1895), p. 185. 

||| Liouville’s Journal (1895), vol. 4, p. 21. 

J] Transactions of the American Mathematical Society (1904), pp. 214-219. 
*** Proceedings of the London Mathematical Society (1904), p. 392. 


M. 0. TRIPP: GROUPS OF ORDER p*q’ 5 


groups of orders p’q, p*q, and p’q’. Ina few cases the existence of the G,. 
requires that one of the primes shall be of a certain form. In § 4 (ii) the exist- 
ence of one G,,,. requires that g = 8n + 3, while the existence of another type 
requires that g= 8n+7. No similar case, in other writings, has come under 


my notice. 


2. Discussion of those G,.,. having neither an invariant H,; nor an invariant H,. 
From Sylow’s theorem we know that if r* is the highest power of a prime r 
which divides the order of a group, the group containsa H.. Hence our G 
must contain one or more [7, and, also, one or more f.. 

(i) p>q. We cannot have ¢H,;, for this requires that q = 1 (mod p) and 
henceq>p. If there are g’H,, then q’ = 1 (mod p) and since we suppose 
p>q we must have 


pq? 


qg=-—1 (mod p). 


Hence we must have p=3,q=2. 
(ii) p<q. Wecannot have pH,,, for then we would have the congruence 


p=1 (mod q) 
and therefore p>q. If there are p’*H_., then we must have the congruence 
p’=1 (mod q) 
which, in view of the hypothesis that p < q, gives 
p=-—1 (mod q). 
Hence PCW Giesia. 
If there are p*H_,, either 
p=1 (mod q) or p*+p+1=0 (mod q). 


The former congruence is impossible, for we suppose p <q. Since we are con- 
sidering the case of q or q’H,;, we must have 


q=kp—1 or q=kp-+1 (k=a positive integer). 


(2) gq=kp—1. 
We must now have the relations 


P+pt+l=lq (1a positive integer ). 
qt1l=kp. 
Since g> p and gq = — 1 (mod p) we must have 
l<p and /=—1 (mod p) 
and, therefore, l=p—1. 


6 M. 0. TRIPP: GROUPS OF ORDER Pe 
Hence p'+p+1=0[mod (p—1)]. 


If p + 2, then p — 1 is always even and p*+ p+ 1 is odd. Hence we must 
have p = 2 and therefore q =7 


(5) q=kp+ 1. 

We also have pt+ptl=_lq. 

Since g > p and qg = 1 (mod p),/=1, so that 
Pee ses Ts 


The only possibilities then that.a G2 may have neither an invariant EH, nor an 
invariant Hs, are for the orders 72, 108, 392 or for the case in which 


DT AN AN etl: 


Now in (6) where p? + p+ 1=q, two of the p* #,, have an H, in common 
which is invariant in the G,... We thus get a factor group I’,,. with p*H, 
and hence only one H;. ‘Therefore, our Gs. has an invariant H,,,. If now 
this H,;, had an invariant H,, it would be invariant in the whole G2. 
Hence the H,, must have qH;. Now q—1=p’ + p and, since p’ + p is not 
divisible by p’, two Fs have in common an Hi, 2) invariant in the if, bo and 
common to all the H,,, and hence invariant in our G,,.. We thus get a factor 
group 1 a with an invariant Hs, since p<q. Hence G2 has an invariant 
FT 2 The latter has an invariant H, , and hence this H,, is invariant in our 

We can now state the important result, viz.: With the possible exceptions 


peg 
Of Gray Grog, Gaon, All Gis ,2 contain either an variant H,, or an invariant H,. 
G...- The H, common to the 8H,, is invariant in the G,,,, corresponding to 


which we have the factor group I’,,. Since the supposed G,,, has 8H,, the I, 
has 8H, and hence I',, has 1H, leading to an invariant H,, in the G,,,. This 
H,, must have 7H,, for if it had an invariant H, this H, would be invariant in 
the G,,,, contrary to hypothesis. Now these 7H, are the only H, in G,,, and 
two of them have in common an H, which is invariant in the H,,. Hence the 
7H, have an H, in common which is invariant in the G,,,. This gives us a 
factor group I’,, which has 1 H,,, corresponding to which our G,,, has an invari- 

This H,,, contains only 1, and hence the latter is invariant in 


ant ye Be 196 
the G,,,. There is then no type of G,,, in our supposed case. 
The treatment of G 


392 
and G,, in the case under consideration, will be given 
under division LV. 


108 


M. O. TRIPP: GROUPS OF ORDER p’q’* 7 


rt 
G 


pi? HAVING AN INVARIANT H, AND MORE THAN ONE Hf7),. 

Note. In the following 7 is an element of order q’, while 7,, 7, are ele- 
ments of order q. 

3. General considerations. If there are qH,,, then H,, must contain an H, 
each element of which is commutative with an H,,. If now T, or T%, accord- 
ing as [, is non-cyclic or cyclic, is such an element and A any non-identical 


element of H,,, while B is a properly chosen element of H_,, then 


and best Wy as) 
Hence Dea Ags) AS as | 
TALL fA 


It follows then that 7, is commutative with each element of an fH, cer Lie che 
H,, is cyclic, then just as above 


AN TTA = 1. 


Since { 7'} is invariant we also have 


ADA == 1. 
Therefore ATTI1A = T% = T%, 
Hence aq=q (mod q’) 
Sane a=kq+1 


and since A is of order p, p* or p*, we must have from the above 
(kg +1)’, (kg +1)” or (kg +1)" =1 (mod q’). 
Each of these three cases requires that £ = 0 (mod q) and hence 
a@=1 (mod q’). 


This makes A and 7’ commutative contrary to the hypothesis that there is more 
than 1H. Hence the case of gH, and an invariant cyclic H, cannot occur. 
If there are q’ H,, we may have either 


q=1 (mod p) or =—1 (mod p). 


If p = 2 these two congruences are identical. 
We will take the different types of H,, and discuss all possible G,... obtained 
with each type. 


8 M. 0. TRIPP: GROUPS OF ORDER p’q’ 


4, H,; cyclic, that is, A? = 1. 
(i) Let there be gH,,. Here 


q=1 (mod p). 
The i, must be non-cyclic (§ 3) and hence 
Are lan 


Besides { 7,} there are q other HH, in our invariant H,. These gH, may be 

divided into 7 sets of p, p” or p* each, the groups of each set being permuted 

cyclically by A; there will remain mH_,, each of which is invariant under 4, 

Hence at least one of these qH_ is invariant under A, so that we may assume 
SL) Ale he 

We may now have three types of G,:,. according as a is a primitive root of one 

of the three following congruences : 


a? =1 (modq), a“=1 (modq), a*=1 (modq). 
Each group thus formed is the direct product of { A, 7,} and {7}. The 


H,; have in common an H/,, H, and H, respectively. 

(ii) Let us take q’ H,, and g =1 (mod p). 

For cyclo H,, ATTA = T°. 
Again we have three types of G,.,. according as a is a primitive root of one of 
the three congruences : 

a? =1 (mod4q’), a” =1 (mod q’), a”*=1 (mod q’). 
Non-cychic H,,. If p> 2, then since 
q=1 (mod p) 

we have q+1=2 (mod p). 


Hence by the same reasoning as in (i) 2H, are each invariant under A. How- 
ever when p= 2 there may be no H_ invariant under A. In every case, if 
there is one H invariant under A there will be at least two. 

We will first consider the case in which p = 2 and there is no H_ invariant 
in G2,2. Hence we may assume 


Amd) Aral: 
and Ar Ae scl 


A? cannot be permutable with 7,, for then {7,7,} would be invariant in G,s,2. 


M. 0. TRIPP: GROUPS OF ORDER p’g’ 9 


It A* is the lowest power of A permutable with 7,, then 
ANT A=f,, 
Apel Al = Arie een) OP 
Ate) Agee Ase LAs heerlen. 
Sh Peed Estos Ter bg Bae boa bea als 
Hence we must have the congruences 
OG. Goes 1 


B(2a 4 BF) = ah ree 


The solution 6 = 0, a = + 1 has already been excluded since it makes 7, per- 
mutable with A’. The solution b= 0, a= W— 1 gives one type of G,.. In 
this G,,. every H, is invariant in an H,,.= { A’, T,, T,}. 

Since we suppose there is no £7, invariant in our G,.., then if x is a Galoisian 
imaginary, we have 
(1) ATTA = (P2785 = Te Te, 


that is, there exists no real number x which will satisfy the above equation 
fan p= 0, 1,2,3;,---,9g—1). Since 


AGEL eAver 1) 
and ah An 
we have Ames Aje=k eh Ss 


Comparing exponents in (1) and above we have 
ac=— 8, Bxe=a (modq), 
whence 2 =—1 (modgq). 


Since x cannot be a real number q must be of the form 4mm + 3. 


Again if ; 2a+6?=0 (mod q) 
and, therefore, b> =—2a (modq). 
We have a=—1 (modq). 
Hence 
(2) ab? = 2a (modq). 
Now 


(3) AGE hatin Avena) sh sauces 


10 M. 0. TRIPP: GROUPS OF ORDER p*q’ 


From (1) and (3) we have 
a8 = an, a+b8=Bfx (mod q) 


and hence we get a’? —be=a (modq), 
whence 
(4) (22 —b)Y = 4(a?— bx) +b? =2a (modq). 


But from (2), 2a is a quadratic remainder and hence real values of # exist which 
will satisfy (4) contrary to the hypothesis that x is not real. Hence the sup- 
position 2a + 6? = 0 (mod q) does not lead to a type of G,s,2. 

Suppose, then, A® is the lowest power of A commutative with any element of 
H,, say T,; and suppose first that 


Ane Maciel, 
that is, 7; is not transformed into a power of itself by A*. Hence 
Ve Beeld We eit Need ME 
Therefore YN Ed Mas Red Be 
If 7, is put in place of 7, T, we will have 
Aya MAS ane 


contrary to hypothesis. It follows, then, that A* transforms 7, into one of its 
powers. ‘Therefore 


hehe A Bella Bin 
Let Aa Aled © 
and a Ratan hette DSS Regt hs 
Hence we must have the congruences 
(5) a+ab*=—1, b(2a + 6?) =0 (modq). 
164 b=0 then a= —1, 


Hence q is of the form 4m + 1 since a is real; a is a primitive root of 
a*=1 (modq). 
This gives one type of G,.. Each H, is invariant in a G,, = {A’, 7,, 7) }. 


Tf + 2a=0, 


— 


then from (5) yh 


But from (4) (22 — b/) = 2a. 


M. O. TRIPP: GROUPS OF ORDER p*q’ 11 


We must now consider two cases according aaa=-+lor—1. If 


a=al, 
then we have the congruences 


P=—2, (22 —b)Y=2 (modq). 


Since x is not real gis of the form 8n+ 38. The former congruence has only 
one pair of roots. That each of these roots furnishes the same type of group 
may be established as follows. Let us take the relations 


a ae RB Ren WR 
and change generators by setting 


cA ear Tals, ibys hes, 


so that our Ge = {A,, T;, T,}- 
Hence AS ek eed AA ES Mo 
ici a=—l, 


we have the congruences 
6? = 2, (2a—by=—2 (modq). 


Hence q is of the form 8n +7. Here again we get a single type of group. 
Suppose 
(6) Jota ee Sew bh then also PAL, abseihaets 


neither a nor 6 can be unity. 

If p> 2 then p cannot divide g + 1 and hence we must inevitably have rela- 
tions of the form (6). We may now have the following cases: a and 6 both 
primitive roots of 


(a) z2?==1 (modq). 
(8) zP°=1, 
(y) ge == 1, 


(5) a a primitive root of z? = 1 while 6 is a primitive root of 2?° = 1. 
(e) a as in (6) while 6 is a primitive root of 2” = 1. 
(§) aa primitive root of 2° =1 with 6 as in (e). 
As regards the three cases (a) (8) (7) relations (6) may be written 
al Ate FART MEd hee (x prime top). 
Transforming with A” in place of A, where y is prime to p, 


AVT AY T#,  AWT Av = To, 


12 M. 0. TRIPP: GROUPS OF ORDER p’¢’ 


If y is so taken that 
ay = 1 (mod p), 1 (mod p’), 1 (mod p*) 


for the three cases respectively, we have the same relations as before with y in 
place of 2 and with 7’, in place of 7,. Hence the number of types is the num- 
ber of solutions of the three above congruences, the solutions (z,, y,) being 
regarded the same as the solution (y,, x,). Therefore in 

case (a) the number of types is (p + 1)/2 for p odd and a single type for p = 2, 
case (8) the number of types is (p? — p + 2)/2 for p odd or even, 

case () the number of types is (p* — p* + 2)/2 for p odd and four types for 
p=2. If x=1 in these three cases, all the (¢ + 1) H, of our invariant H,, 
are invariant in the whole G,,,..__ In case (6) there are p — 1 types, since b may 
be fixed as any one of the primitive roots of x”*=1 (mod q) and there are p—1 
types corresponding to the p—1 values of a. In like manner case (e) furnishes 
p — 1 types and case (£)p(p — 1) types. 


5. gH, and g=—1(modp). Here we will take p> 2, for the case p = 2 
has already been treated in §4. A cannot transform 7, into one of its powers, 


for if we had 
a een oe 


then since a = 1(mod ¢) it would have to be a primitive root of x, x* or 
«**=1(mod q). Clearly this is impossible. We must therefore have the 
relations : 


Agel Alsi, Ag Al Tien. 
Using Galoisian imaginaries we may write 
A ial Arsh iiss 


Proceeding as BuRNSIDE does in his Theory of Groups, pp. 136-7, we see 
that there are three types of groups in our case according as 7 belongs to the 
exponent p, exponent p? or exponent p* (mod q), that is, according as A”, A?’ 
or A” is the lowest power of A permutable with Z,. The second and third 
cases require g=—1(modp’) and g=—1(modp’) respectively. Here 
a=—1landb=i+i1(modq). 

I give the following illustration of the use of Galoisian imaginaries for find- 
ing 6 in the case where 7 is a primitive root of the congruence 


of ==) 1 (m0. 7). 


Let us take a G,.= Gis... Since 2 is quadratic non-remainder (mod 19) we 
form the irreducible function 
F(a2) =a? — 2. 


M. O. TRIPP: GROUPS OF ORDER p’*q’ 13 


Cf. HOLpER, Mathematische Annalen, vol. 43, pp. 350-1; also Dicx- 
son, Linear Groups, § 6. 

We must now find a function f(), i. e., a mark of our Galois field (ef. 
Dickson, Linear Groups, p. 7), such that 


[f(v)]>=1 {mod [19, 2? —2]}. 


The period of our mark is 5 and is a divisor of 19? — 1 (cf. Dickson, loc. cit., 
p- 11). To obtain this mark we proceed by trial. 
(1) The different powers of x mod [19, x? — 2] are 


Ae eee nO als eo eo sr LO EL 2 
@, 2, 20, 4, 4v, 8, 8a, 16, 162, —6, — 62, 7, 
Io alae Ome elie) 186 
in Laveen oy) Coes) a Li 





Hence a®=1 {mod[19, a—2]}, 


that is, the mark x belongs to the exponent 36 and since 36 is not divisible by 
5, none of the powers of x given above can be taken as our mark of period 5. 
(2) Let us now try the powers of 1 + # mod [19, 2? —2], 





1, 2, 3, 4, D, 6, ts 8, 
l+a, 3422, 5a+7, —2—7x, 34107, 18~a+4, 22+11, 974+7, 
9, 10, ati 20, rosy 40, 
meerGesr, 9 \-.., | —1, ee eri 
Hence _[lt+-2)]'=1 {mod [19, 2’—2]}, 
that is (97+7)P=1 {mod [19, 2?—2]}. 
We take t=9r+7 {mod [19, 2?— 2]}, 


and hence i?=i%=(7+492)'=7+102 {mod [19, 2?—2]}, 
and b=it+%=1974+14=14. 


Hence our G is defined by the relations : 


5°. 19? 
tied A TdT AP TATU TTT. 


The result obtained above may be verified as follows. Let us represent an 
isomorphism of { 7,, Z,} which is invariant in the whole group by 


ee 
T=(n o2'm) 


14 M. 0. TRIPP: GROUPS OF ORDER p*q’ 


This is the transformation of { Z,, 7} under the element A. 
Ae -( an uel cael i) 
Te Tae 
Je = ( Pes ae ea) ’ 
it ae ? 14: th 
Rese 7; a; 
= TON) Tease) 1 bes ent v hspeeh ’ 


ay Sis f,, yi 
oe Tey eae hee Grats Ered ec . 


Since J° is here the identical isomorphism, we must have the two congruences 
b§— 36°+1 = O(mod 19), —b?+2b=1 (mod 19). 
By trial we find that these two congruences are satisfied by 6 = 14, and from 
the method of forming these isomorphisms it is evident that 5 = 14 will also 
satisfy the congruence 
b° — 467+ 3b =1 (mod 19). 

6. Va hea shots blk We sees sl SUD 2 ote 

(i) Let there be gH. Therefore g=1(modp). The H,, is non-cyclic 
(§ 3) and hence we may assume the relations : 
A> DA= 1), BULB = Ty, (Ayia 
Therefore deans bet hal a, WEN eLOd Me ta hf bi 
and Avid, Lt, Ate ee as 
Since AB = BA, we have the congruence 

1+aB=1+8. 

Hence a=l1 or B=0. 


In the former case A is permutable with each of the q + 1H, of H,, and as B 
is permutable with two of them, there exists at least 2H, invariant in the whole 


Gs 2° Hence we have the relations: 


AT Ajeet Tig BTA TS SAAT Aor) Ane Bn 


If a =1 (mod q) and 6 belongs to the exponent p (mod q) we get one G,,. 
which is the direct product of {7,, A} and {B, 7,}. The q#,; have in com- 
mon an H,,= {A}. If 6=1 our G,,, is the direct product of {B, 7} and 


M. 0. TRIPP: GROUPS OF ORDER p’¢’ 15 


{A, 7}. Here we get two types according as a belongs to the exponent p 
or exponent p’ (mod q). In the former case the H,; have in common an 
Fi, = { A’, B}, while in the latter case they have in common an H, = { B}. 

If a and 6 belong to the exponent p (mod q) we set 6 = aY and we can then 
put in place of A, A, = A*B (2 prime to p), keeping B fixed, and therefore 

Me oACe Tapas 1 As. ade 
If @ is so chosen that 
rx+y=0 


this case reduces to one in which a=1. Again if a belongs to the exponent 
p’ and b to the exponent p (mod q) we may set a? =b*. We next keep A 
fixed and in place of B put B, = A’ B’ (r prime top). Hence 
Tap A dap mo dag. Neral ie bad ogee beat 
If 7 is so chosen that 
z+r=0 


this case reduces to one in which b=1. 
(ii) Suppose there are q’ #7, and also g = 1 (mod p). 
For cyclic H,, we have the relations 
Age LA aa Es, BA TB =.T”. 


Just as in the preceding we get three types of groups. Two of them are the 
direct products of { B} and {7, A}; the other is the direct product of { A } 
and {7, B}. 

If p> 2, then for H,, non-cyclic we may always assume the relations : 
feted Bb Tt ATT A= Ts, BOL Ba T). 
For suppose there is only 1/7,, say { 7,} invariant in our G,s,, then 
eee ae 2, BOT Ba TT, AXTA= 73, BUT, B= Tih. 
Hence 

Pte LA be OR er = An Bo 2, BA ae Te SY 
Therefore 8 = 0 or a = a and accordingly relations (6) reduce to (a) just as in 


(i) above. Again suppose there is no #7, invariant in G,... Then we have 
relations : 


Paela BOT R= Ts, AOT.A=T*, BOT,B=T*Ty. 
Now BAT TIA soya et ey, 
and AOBIT TBA = Parse Teter, 


whence 8 = 0 or a =a and accordingly relations (c) reduce to (a). 


16 M. O. TRIPP: GROUPS OF ORDER p’q’ 


In (a) we cannot have «= 8 =1 or a=b =1, for then there would be only 
qi,;. Just as in (i) above we can make one of the exponents a, 6, 2, B=1. 

(1) If2=b=1 we get two types according as a belongs to the exponent p 
or exponent p” (mod q). 

(2) Ifa=a =1 we get(p + 1)/2 types, each being the direct product of {A } 
and {B, 7,, T,}. With regard to the last named subgroup see HOLDER, 
Mathematische Annalen, Vol. 43, pp. 341-45. 

(3) If b= 8=1 each G,,, is the direct product of {B} and { A, 7, 7,} 
and, therefore, we have three cases (cf. LEVAVASSEUR, I. c., pp. 339-41). 

(a) Ifa and a both belong to the exponent p (mod q) there are (p +1)/2 types. 

(8) Ifaand a both belong to the exponent p’ (mod q) there are (p? — p +2)/2 
types. 

(y) If a belongs to the exponent p while a belongs to the exponent p? (mod q), 
there are p — 1 types. 

(4) If «=1; a,b, 8 = 1 and a belongs to the exponent p (mod q) we get 
(p+ 1)/2 types. If a belongs to the exponent p’, this case reduces to one of 
the preceding. 

(5) If B=1; a, a, b = 1 then to get new types we must have a belonging 
to the exponent p? (mod q). Ifa belongs to the exponent p’ (mod q) we get 
(p? — p + 2)/2 types, while if « belongs to the exponent p (mod q) we get 
p —1 types. 

If p= 2, B must transform some element of H,, say 7, into one of its 
powers; and, therefore, a second element, say 7,, into one of its powers also. 


Hence Baw Dias los BOT Baaelee 

If we also have Aglaia Ls 

then we can apply treatment similar to the above for p > 2.. But if 
As Ti Ape= ol 

and hence Agi A late, 


then proceeding just as in § 4, where p= 2 and A‘ is the lowest power of A 
permutable with 7,, we find b=0anda=—1. If a=fB=+1 we get one 
type of G,. which is the direct product of {B} and {A, 7,, 7,}. Asin §4 
q must be of the form 4m +3. There can be no other type with the relations 


AAT ART, BIR Bags CAST AT her 
For BA4I1T TAB = TET 
and A“BT TBA = T*T=8, 


M. O. TRIPP: GROUPS OF ORDER p’¢’ 17 


Hence BS=a=+1. 
8 = —1, we can keep A fixed and set B, = A’B in 


If, however, we take a 
place of B so that 
Bik, Dot be Aa Ar bi=sl 4 


Hence the value — 1 for a and B gives the same type as + 1. 


7. @H,;, and q=—1 (mod p)(p>2). Neither A nor B can transform 
any element of our non-cyclic /7,., say 7,, into one of its powers different from 
unity. For if we had 


Jil AAAs a eet Es or Beal Death (a,b+#1) 
we would have g=1 (mod p) which is impossible. We may have the relations: 
feel A= 7. Jalen be os bee BV AM, ial hs IB Wigley ee Sls 


These give two types of G,:,., each being the direct product of {A, 7,, Z,} 
and { B}. The exponents a and 6 are determined as in § 5. 
We may also have the relations : 


ee A ea Te RIT BHT! BT Ae pre 


aand # satisfying the same relations as a and 6 for the corresponding case in 
$5. This gives one type of G,,,., the direct product of {A} and {B, 7), 7;,}. 

The hypothesis that no element of {A, B} is permutable with an H, of 
{T,, T,} is inadmissible. This follows from the discussion of the isomorphisms 
of the non-cyclic H,, by LEVAVASSEUR in his Enumération des Groupes @ Oper- 
ations d’ Ordre donnée, p. 52. The conclusion of this discussion is stated in his 
article on G‘,2,2, loc. cit., p. 349, as follows : ‘“‘The substitutions with character- 
istic irreducible congruences * divide into cyclic groups J + forming a complete 
and unique series of conjugate subgroups ; then two isomorphisms correspond- 
ing to two such substitutions can only be permutable if the one is a power of 
the other.” From the above it is evident that no type of G,,,. with cyclic 
#,, exists in our supposed case. 


ee Ae OP I AB BA AC CA, BCO— CA}, 


(i) Suppose there are g/Z,; and hence g=1(modp). The /,, is non-cyclic 
and the most general form of our relations is 


A“TA=T7, BT.B=T, OT. C=T, 
Mee Peten Tel HoT De GAT OT Oli Ts: 


*Cf. HOLDER, Mathematische Annalen, vol. 43, pp. 348-9; also BURNSIDE, Theory of 
Groups, p. 136. 
} See 25 for this J. 


18 M. O. TRIPP: GROUPS OF ORDER p’q’ 


On ue TT, by AB= BA as in §6 we see that we may assume 
s= Next on transforming 7,7, with AC = CA we have either a = 1 or 
y= Af If a=1 we then transform 7) 7, with BC = CB and thus find that 
b=1 


or y = 0, so that we may assume the relations : 
AT A= 1; Das aici, Cine sand. 
Ts bate BV R= he Bil Beek OT eee 


If a, 6, ¢ = 1 then by the same process as in § 6 where a and b both belong 
to the exponent p (mod gq) we can make a=1. Then keeping A fixed we can 
repeat the process so as to make b=1. Hence we may assume a=b=1 while 
e must belong to the exponent p (mod q), so that we get one type of G,s,. which 
is the direct product of {7,} {A}{B} and {C, 7}. The #, have in com- 
mon an IT) = ele ns 

(ii) Suppose there are q’ #7; and also g=1 (mod p). If the ZH, is cyclic 
we have the relations 

Ve leas Ge" BE) A Dal DB eal Orel Creare 
By proper change of generators, just as above, we can assume a = b =1 and 
hence we get a single type. 

For #7, non-cyclic we make the following suppositions. 


(1) Let the 27, contain at least 2H, with each of which A, B and C are 
commutative. Hence we have 
Ahh, Aa the Veter Mil eee Shy by Chapt kel OV bi 
Ad Areas Bl. Deere Or 1, C= 
As above we can make two of the exponents a, b,c=1, say aandb. Then 
after that we can make one of the exponents a or 8=1. The above relations 
then take the form 
CATS Acer e Bohl, Cor Oa ehe 
Fe keayy bo. byes of A Ba eget, Cr, Cairo 
Hence these groups are always decomposable. If 8 =1 we get (p + 1)/2 types 
[c.f. § 4 (ii), case (a)] which are the direct products of { A, B} and {C, 7,, 
T,}. If y=1, 8 +1 we get a type of G,,,. which is the direct product of 
, C} and {7,, B}. If c, 8, ¥, are different from unity we can, by a 
proper change of generators, reduce to one of the preceding cases. 
(2) Suppose there is only one #, invariant in the whole G,.. Let this 27, 
be generated by Z,. We may take the element A commutative with { Z) } and, 
therefore, we have the relations 


Anil Ae Ts eRe Bae COU HOES TE: 
ATA S Te) BAT BR rs OAT eee 


M. 0. TRIPP: GROUPS OF ORDER p'q’ 19 


We now transform and make use of the fact that AB = BA. 
DEAT VMAS ES VER EG Kes td Mae tindd Hh 
and A?BOT TBA = AIT TBA = TH+" Te, 


Hence /=0 or a=a. In the latter case A would be commutative with all 
the (q+ 1), and as B is commutative with two of them, we may now write 
our relations in the form 

Be Alms his ales as, ee ML OR bids 

TNT AA eet of bl SEAS bi oy Oe Me Op Wf EP Moat bs 
Transform, making use of AC= CA just as above. Then a =a, since m=0 
makes 2/7, invariant in G,,,. contrary to hypothesis. Again transform, using 
BC=CB. This makes }=8. Since a=a and b= 8 all the vif: are per- 
mutable with A and B, and since 24 are permutable with C, we find that the 
case of only one /7, invariant in G;,. is impossible. 

(3) Suppose there is no /Z, invariant in our G,.. On account of the 


congruence 
qg=1 (mod p) 


2H, must be commutative with A, 2H, commutative with B, and 2H, commu- 
tative with C. Our relations may now be written 
ah A em: TS, Breen ye, Cant san le de, 
Fler A BL Se eel Met A gs A Spd Sd shy Ae ae CO eM AA ESET ad de 
Let us transform as follows : 


Bers Be a oS COTE Sad Md ALAD 3 RI a Aiea 
and Deh ied bi ie UE Bs Meee Moa MEY Moll Ak bk 
Hence a5 = 0, so that either a = 0 or 6 =0, and proceeding as in (2) above 
we see that our relations may be written in the form 
ate Ae 13 ls bed ag os dbo Cir AN OR Td Heed | 
ieee aie= "055 Las aa OO TO aires. 
Again let us transform thus, 
Jeet NO ea WENO EY SS Te tad MEY had 5 EEO he py do 


and C7 BT, BO= CT? C= T8" T8, 


Whence b = 8, since y + 0; for otherwise 1/7, would be invariant in the whole 
group. Using A and C we can showa=a. Hence just as in (2) hypothesis 
(3) is absurd. 


20 M. 0. TRIPP: GROUPS OF ORDER p’q’ 


9. @ H, and g=—1(modp)(p>2). We may have one type of Gs, 
defined by the relations : 


A“TA=T,, B'T, B=T,, C*T,C=T, 
ADT A TUTE NB TAB ee TC a TA nr 


6 =i -- i where 7 is a Galoisian imaginary. This G,;,. is the direct product of 
{A, T,, T,} and {B, C}. There can be no other type in this case (§ 7). 


10. HH, of the type A‘ = BP=1, BT AB= A’. 
(i) Suppose there are gH,. Hence the H,, are non-cyclic. Therefore 
($ 3) Velen 18 ches Sis Be Daas 


Since B must be permutable with a second £,, say { Z,} we must have the 
relations 


Da writ bah fh, AG Ara he 
We transform making use of the relation AB = BA*. Hence 
BRAT TAB =) Tie AS BAT Tp Ase oto 
Therefore we must have the congruences : 
bB =b'B, a=fPa(1+6+06’) (modq). 


8 must have the values +1. 
If 8=1 all the H, are commutative with B and hence we may assume 
Shp FEST he 
If 8=—1 thna=—a(1+6+40’). 
If a=0 the last congruence is satisfied and, therefore 


aT Ape Te 


Suppose a= 0 then we must have 6=-+-1. Hence we have two cases 
according asb=+1orb=—1. Therefore 


1=—1(1+1+1) (modg), 
=—1(1—1+1) (modq). 


or 


Each of these congruences requires that g is divisible by 2, which is absurd. 
Our relations may now be written 


ATCA Te Teme yee 
ART Ae [ey Oh UTD are 


where band 8 take the values +1, but both cannot be +1 under our supposition. 


M. 0. TRIPP: GROUPS OF ORDER p’q’ 21 


If =—1, B=+1 we get one type of G,.. 
LE, b=+1, B=—1 we get another type of G,... 
b=—1, B=—1 does not give a new type, 


for AB in place of B leaves 7, fixed. Both these types of G,. are the direct 
products of {7} and {7,, A, B}. 
(iil) @’H,. For the cyclic 1, we have the relations, 


ATA = T*, ISM sie HAE 
Since AB = BA? we transform as follows; 


BTADSTAB = T° = ASBITBA = TT”. 


Hence a=+t. Also J6=+1. 

If a=+1 and b = —1 we get a G,, in which { A } is invariant. 

If a=—1andb= +1 we get a G,, in which { A} is not invariant. 
a=—1, b= -—1 gives the same type as the latter, 


for we may take AB in place of B and then b= +1. 
Non-cyclic H,. 
(1) Suppose that 47, has 2H, invariant in G,,. Therefore 


oe RT Rea AIT AT: | BOT RTS. 


The same procedure as above shows thata,a=-+1. We accordingly get four 
types with the following relations : 


PeewaA 7 RIT RT, AIT A= TS, BAT B= Ty. 
ee eA BT Be TO ADT A Tl BOT Bo Te, 
Reef) eB TB Ts) AOA Tt) OBIT Ra To 
CEA eB T RT AT Awe Te BA TOR To. 


(2) Suppose only one / is invariant in the whole G,,. We may now write 
the relations 


Seater ee ei) ADA 78) BOI BD = TP TY. 
As in the preceding case a? =1. Also 

Yep walk AU Deja Ahem Id ke) 
and a lead agen KSA Gages A Movant as W bia 


Hence 8 = 0 or a = a and, therefore this case is impossible. 
(3) Suppose no / is invariant in our G,.. 


22 M. O. TRIPP: GROUPS OF ORDER p*q’ 
(a) If there is one HZ, and, therefore, two H, invariant under A we have 
BAL A bere hay Bape a em ied eae 
VePara therein thos, BAT, Bes TOE, 
Transforming Ma gare Wie Meg Mies Wa Pe Fad koyroik dd bce at | 
and Me baad 6 Sri hel AAW a ie Rd Abstain sh AI Mec! OR 
Hence we get the four congruences : 
ab =a'b, aB=a'P, ac = a’c, ay = ay. 


If 6 + 0, then a? =1; and since 8 = 0, we have a=a. But if a=a all 
the H, are invariant under A and since 2/7 are invariant under B we have 
case (1). Hence we must assume 6=0.. Likewise we must havey=0O. It 
is also seen that a=a* and a= a’, a and @ belonging to the exponent 4 
(mod q). Let us now put 7’) in place of 73 and keep 7; fixed. Therefore 


Bape MT 64 vot ke. 
and TB gat tM a eae 0 iis SAL ore fi 3 
But Dies BT BS) m5 de ed ee 
whence Be =1 (modq). 


Hence, dropping primes, we may write our relations : 
Ad Ale ed en Baylies As) 1) Alesis BT, Baas 


No matter how we transform or change our generators no inconsistency in these 
relations arises, and hence we get a single type of G,.. 

(b) If no Z, is invariant under A, proceeding as in § 3 for the corresponding 
case, we have the relations ; 


ANP ALT, BATRA) NA A= 7 i 7 er 


Therefore BAGEL | AD =iisemh : eel eel 
and Plea oraicrth W eee hice we hiragy egos ey Fo 
Hence we have a = 0 and 8 = —a. This gives but a single type of G,., for 
a=+1,8=-—1 and a=—1, 8 =+1 merely amount to an interchange 


of 7, and 7,. 


M. 0. TRIPP: GROUPS OF ORDER p'*¢’ 23 
11. H,={At=1, B= A’, B'AB= A}. 


(i) Suppose there are g/7,. In this case the H,, are non-cyclic. Our rela- 
tions may now be written in the form 


eee eA eet han 7 CANT AS Ts) BIT BR TeT*. 
Transforming 

me An 1 AB a TP Te and DONA Epp at hd a 5 ENT A EY Bal) 
Since AB = BA* we see that 8 = 0 or a= 1, and hence in either case we 
may write, assuming 8 = 0, 


Fad Ale bees of Me Phe DeLee 
Therefore 
Bae Deas Poa ae el hea 
Hence a’ = 0° and since b = 0, we have a’ = 1 from (2). Ifa,b = —1, by 
taking AB in place of A we find that 7, is transformed into itself; while if 
a=—1,b=+1, by taking A’B in place of A, 7, is transformed into 
itself. Hence, in each case, we can assume a = + 1 and then the only value 6 
can have is —1. It follows, then, that relations (1) give only one type which 
is the direct product of { Z7,} and { 7}, A, B}. 
(ii) q?H,. For the cyclic H, we get, just as in the preceding case, one 
type with the relations : 


ve bea EP OS bag boy tS Ee hg 


where a = + 1 and 6b = — 1 (mod qq’). 
Non-cyclic H,.. 


(1) Suppose the 7, contains 2/7, invariant in our G,,.. Therefore 
GE UES Meas 3 ed 8 8 Nas de a Ae LS BTA eis Let 


As in (i) we can assume a = +1,6=—1(modq). We cannot have a and 
8 both congruent to +1. This gives one type of G,. with the relations 


ee A = 7) Daal tiie”, wis We hc Ned ES Bgeli Dal. 
Only one other type of G,,. is possible, for if we have 

eet A=, Priya ees ge DA ali, Ta A Nad ea Ne 
then by keeping A fixed and replacing B by AB we have 

AS Ms SE AA PEP Te PE) fe el. Ae Lee, de ees 


(2) The case of only one 77, invariant in the whole G,,, may be shown impos- 
sible just as in § 10. 


24 M. 0. TRIPP: GROUPS OF ORDER p*q’ 


(3) Suppose there is no #7, invariant in the whole G,,, but let there be one 
fT, commutative with A. Our relations may now be written 


AMPA = Te) BAT RTT, CAS TOA ae 7 ee 


Transforming AT BT Bs Aree ae eee oe 
and DA Sl Oe AD aaa) Coe are) caaralss 
Hence we have the congruences : 
(1) ab=ad, (2) aB=a'°B, 
(3) ac = ac, (4) ay = aby. 
Also we get by transformation 
I Etat LAW 8 ig asd bh Od Bh ergs ee ANE RS AL 
os DT Dee en eY DAA a A ee ie 
Therefore we have the congruences : 
(5) P+Be=a’, (6) be + yo = 0, 
(7) 8b+Py=0, (8) B+y=e’. 


We must now investigate these eight congruences to determine the values of 
the exponents a, b,c, a, 8, vy. It may be noted that c, 8 + 0 for otherwise 
we have a preceding case. From (7) if 6=0 then y = 0, and if 6+ 0 then 
y += 0, also 6? = 7 and hence from (5) and (8) a’ = a’. 


First, suppose b+0. 
From (1) v=1 
and since by (2) 2=a’ 
we have A=a. 


We may now have two cases. 

(i) B is permutable with one /7, and, therefore, with two ZZ,. This is 
clearly the same as case (1) since 2/7, are invariant in G, ». 

(ii) B is permutable with no ZH. Now 


B-! the fh liz Sigs ti ha i = o big CM byes. J 
by (7). Taking ¢ as a Galoisian imaginary we have 


BOT? T! B= (T?T1)°. 


M. O. TRIPP: GROUPS OF ORDER PF 25 


Hence we must have the congruences 


ba + By = ox, 


ca + by = oy. 
Eliminating x and y we have 


c=c8+W=a’ by (5). 


Hence o”* = 1, so that o is real, contrary to hypothesis. 
Secondly, suppose b= 0. Since b, y = 0 we have 


aa, A= 1, DAC ANTE D hing age Cle Daal Dewi 
Let 7) = 7; and then our invariant ITs Sabet |) seme tence 
Je IA ayes Peet Kee seemed Hb EM AG 
from (5). 
(a) 2 = +1. Here a = a from (2) so that dropping primes we have 
ee a 2s, Ved bv epee dk Ad [rs BERS big Dae le Desse le. 


These relations show that { 7,7} is invariant in G,,., contrary to hypothesis. 
(b) a = —1. From (2) «2 = —as0 that we have the relations 


eee eh Re TAT ALT! | BAT Ba 7, 


which furnish a single type of G,,.. 

(4) Suppose no non-identical element of {A, B} is commutative with an 

7 Lhen each non-identical element of {4} and of { B} corresponds to a 
non-identical isomorphism of /7,,. According to § 7 the substitutions or iso- 
morphisms of . with irreducible congruences divide into groups J forming a 
single conjugate set. Since {A} and { B} are not conjugate the correspond- 
ing groups of substitutions are not conjugate, and thus we have a contradiction. 
Accordingly there is no type of G',,. in the present case. As regards the 
correspondence between the elements of {.4, 6} and the isomorphisms J, see 
Ho6iprer, Mathematische Annalen, Vol. 43, p. 329. 


12. digest | AS He Peel, Deed pies A) ( p Odd) 
P 
(i) gi. Since the 7, must be non-cyclic we have the relations ; 
Crm ae Be Tbr AR A al BAT, Bie De, 
for if we have B-! 7, B= TT} instead of the last relation above, then 
De AIRY RA WA Os os at RR Ei: 


and APA BATT. BAM a= Tite Pa 


26 M. 0. TRIPP: GROUPS OF ORDER p*q’ 


Therefore c = 0 or a = 1 and hence in either case we have relations of the form 
(a). From the above transformation we have 


a’? =1 (mod q). 


If a, b = 1 (mod qg) we can choose an integer k such that ab* = 1, and if we 
take A#B* as a generator in place of A, keeping B tixed, T, is transformed into 
itself. Hence we may assume a = 1 and our relations become 


AAT Ame Ty BAT R eiT | ACME A eT rr 


where b belongs to the exponent p (mod g). Any G,,. formed from these 
relations is the direct product of { 7,} and { 7,,4,B}. WHsTERN (l.c., p. 223) 
shows that there are p — 1 types of {7,, A, B} corresponding to the p—1 
primitive roots of 

a? =1 (modq). 
Hence we have p — 1 types of G,,,2. 

If 6 =1, a= 1 in relations (a) then a belongs to the exponent p (mod q) 
and we get one type of G,:,., for taking A* in place of A all our relations are 
unaltered, except that a is replaced by a’. 

(ii) Let there be q’?H; and also g=1(modp). For cyclic H, our rela- 
tions may be written 

aah TA pe 1%, BO TB = 7". 
If a = 1 and 6 belongs to the exponent p (mod q’), just as in (i) we get p — 1 
types corresponding to the p — 1 primitive roots of 
b?=1 (modq’). 

The same p — | types will be obtained if both a and 6 belong to the exponent 
p (mod q’). If 6 =1 and a belongs to the exponent p (mod q’) we have one 
type. Just as in (i) it may be shown that @ cannot belong to the exponent 
p* (mod q*). 

Non-cyclic IT. 

(1) Suppose 2H) are invariant in the whole G,,,.. Our relations may now 
be written 

CA Ae Te psn Biel oI bee Ve het Ali byes AC BUT, B =e 
Neither a nor a can belong to the exponent p’?(modq). If a, b = 1 we can 
change generators, as above in (i), so as to make a =1. Hence we have two 
cases : 

(I) a = 1 (mod q) b belonging to exponent p (mod q). 
(II) 6 = 1 (mod q) a belonging to exponent p (mod q). 
If in (1) a = 1, we have the relations : 


AD A oe Di.) BOT eT), AGT A ee Ce Bake ele 


M. 0. TRIPP: GROUPS OF ORDER pq’ 27 


Since different primitive roots b, 8 furnish different types, we have here 
(p — 1) types of G:.2. Again, if in (I) «41,8 =1 we have p—1 types 
with the relations : 

mel A 1, Fa ee hae AShTy Aiea: T, Dla 
In case (I) if a, 8 = 1 we have (p — 1)? types with the relations : 

etd == 1, Dele Death Hi bea nee iid Heals Peg Ne Re See 
In case (II) there are 4(p + 1) types [ef. § 6 (ii) (2)] with the relations : 

ert A Jo rel ag = By Aga Als dian U8 srt iid EM SEL 
The case in which a, a, § all belong to the exponent p (mod q) easily reduces 
to a preceding case. 


(2) Suppose there is only one H, invariant in the whole G’,,,.. Our relations 
may now be written | 


era Bo RR Te. | ALTA Te, BIT BR Te Ty. 


As in the preceding discussion we may assume that a belongs to the exponent 
p(modq). By transformation 


eae eh A. Dea) sh: 714 RAH ORNL ESN 8 Eee EE VA Rapes tH Me 
Hence aB =a. 


If 8 = 0 we have two H, invariant in G,.... Hence 8 + 0 and therefore 
aza. It follows, then, that all the H, are invariant under A and hence we 
may pick out our generators so that two H, are invariant in the whole group, 
contrary to hypothesis. 

(3) Suppose there is no H, invariant in the G,,,.. Since there will always 
be two H, invariant under A, our relations may be written 


Pee ed Dima tT oy a A 8, BT B= Te TY. 
Peicaosiormation BOAT Ts 7! AB as Tetoby Posto 
and A-(etvU Bat fe Ty BAvtt xt TaP bat aP By Y Ma aaa } 


We must have c, 8 + 0 and hence, comparing exponents, 


a = a? and a = art, 
ae yPt+l — p+1\p+1 — p,p?t+-2pt+1 — »2p+l 
Hence as wt = (atl) = « = o'Ptl . 
Whence fe gee 08 | 


and therefore e+]. 


28 M. O. TRIPP: GROUPS OF ORDER p*q’ 


Since p is odd we cannot have a? = — 1 so that 


Hence a = a and since all the H, are invariant under A and two H, are 
invariant under B, two H, are invariant in the whole group, contrary to 
hypothesis. 


13. ¢ H,; and ¢q = —1(mod p). No element of order ¢g can be trans- 
formed by A or B into a power of itself different from unity. Let us consider 
the relations : 


1) AST Ae Bee ae ee lee BOT Besa 
1 2 1 1 2 Dies 207 2 2 


Since Janel dhe Vane 

then AG BAG Ano AG 

and since A" BA= BA? 

we have BAG =e eel 
Therefore (Ae Aerts 

Again Agata 1 A) Abe Are (Ol ees ae 
Hence gal Maren ts irs A Re I Bi A 

and TT?) = A?Tz1A?. 


It follows then that A?’ and 7,, 7, are commutative. Hence relations (1) 
furnish but a single type of G,,., where 6 is determined by Galoisian imagi- 
naries as in § 3 (ii). 

Next we will consider the relations 


(QA TA = TBAT OB Sa 71o aa), Ala eee 
The relation B-'T, B = T, can be written B-'T, B = T‘ where i is a Galoisian 
imaginary and a primitive root of the congruence 

#”=1 (mod gq). 


Different primitive roots of this congruence give different types of groups (ef. 
WESTERN, |. c., p. 223). Hence relations (2) give p — 1 types of G,s,. 

There is no further type, for the supposition that no element of { A, B} is 
commutative with an H) is inadmissible, just as in § 11 for the corresponding 
case. 


M. 0. TRIPP: GROUPS OF ORDER pq’ 29 


14. Hs = { A? = B= C?=1,AB=BA,AC= CA, By CB = Aa Ct 


(p odd). 
(i) Suppose there are g7,. Hence 


q=1 (mod p). 
Since the H_, is non-cyclic our relations may be written in the form 
EIB ts od be Pa cI Nagen tls Coal 
) FSS NT SARE TG 9 ce IEF od Mad Bae OF TEG = 1773), 
Now Diesel PAL paid ce 1 o8 
and Acie mild is Ake kee Dog 
Hence 8 = 0 or a = 1, and therefore we may assume 8 = 0 in (1). Also 
Co Dell BO Uae, Oz tv Te 
and pA Og el, Cae taa b 1 AD eet he’, 
Since BC = CAB we have the congruences : 
l+qb=1+y and cb = abe. 
Now c # 0 for otherwise 7, would be independent of 7,. Hencea= 1. 


From 
yb=y¥ either y=0 or b=1. 


In the latter case all the H, are permutable with A and B and since 2H7 are 
permutable with C we may assume y = 0, and hence we have the relations : 


ATA? eB WRT, On LOT, 
Amel Ail), SERS EM a Hes IEE BE SINE SW Bh 
If 6, c+ 1 we can set b = c’ and now in place of B let us take BC". Hence 


(2) 


CBT BOt Te. 


If, then, & is so chosen that « + k = 0, 7, is transformed into itself, and accord- 
ingly we can assume 6 = 1 in relations (2). We thus get a single type of 
G 3,2 Which is the direct product of { 7,} and { B, C, A, 7}. 

(ii) Suppose there are q’ 7, with g=1(modp). As in the preceding dis- 
cussion we may write our relations for the cyclic Jef Ap 


ea Alan 7, Boh Bees ly Ore KORY thee 


This gives us a single type of G,., ¢ belonging to the exponent p (mod q’). 


p*q? ) 


30 M. O. TRIPP: GROUPS OF ORDER p’q’ 


For non-cyclic H,, we have different cases : 


(1) Suppose two 7, { T,}, { 7} are invariant in the whole G,,.. Hence 
(a) in ee Bee oles Vi Genelia 
Behe biG hke sii. BO Bie Te OG eaihyy 


As in the preceding we may assume d =a=1 (mod q). We may then 
choose our elements so as to make b= 1 orc=1. Let us say that 


b=1, 


then if 8, y = 1 we may let y = #" and keeping B fixed, put B*C in place of 
C.. Therefore tC Bip Cer, 
Now let & be so chosen that k + y=0. Hence we have the relations 


ALLA PM BUT BT eh aati 
AT Ace TON BAMTER ae TERN gC 


We, thus, have a single type of G,,..; for if we take B, = BY, C, = C*, A, = A” 
the relations for our H/, are unaltered. c¢ and 8 belong to the exponent p 
(mod q). 

If in relations (2) 8 = 1 we get (p + 1)/2 types with the relations 


AMT AT BOAT PB cont TC eee 
AYT Ace Tye) BAT ROME aie teenies 


(2) Suppose only one H, invariant in the whole G,,.. After a proper 
change of generators our relations may be written 


ANT Ame TEBE Ra) AGT Cas ae 
AAT A=T, | BOT R= PM, C77 CL mr, 


Hence Co Ls Os Gente ont ie 
and BAO aid, dy CA ee MGB a lene 
Since Xr, H#0 we havea=1. 


Consequently by a proper change of generators, we may assume 8 = 0. Hence 
AY = ¥ and, since 

y #0; 
we have A=1. 


This makes all the H, invariant under A and B and, since two are invariant 
under C, two H, will be invariant in the whole G,;,., contrary to hypothesis. 


M. 0. TRIPP: GROUPS OF ORDER p’q? 31 


(3) Suppose there is no #7, invariant in the whole G‘,,... Since q=1 
(mod p) and p is odd two /7, at least, are permutable with A or Bor C. In 
{A, B, T,, T,} two H, are invariant under both A and B [ef. § 6 (ii)]. 
Hence we have the relations 

Ae A eel: Pore Led gr He, ese A EE ball Mae 


eA Tea TB Ta ROT Ory Te 


Now Ca Age AG eer ype 

and eG era Ara ote 
Since ay 0 wehave a=a. 

Also Ce Bal DG ey One Tost ey 
and age Ae Ce CA Ls sean Oren crys Yoh art ak ey 


Hence we have the four congruences : 
(1) ch=abe, (2) Bysaby, (3) db=aBd, (4) 88 = a8. 


From (2) we see that if a = 1 then 8 = J, and hence there would be two H, 
invariant in the whole G,,.. Therefore we must havea, a1. Keeping A 
and C’ fixed and taking A*B in place of B we can assume b = 1, provided k is 
properly chosen. 

From (2) y8 = aby = ay and since y #0, S=a. 

From (4) 85 = a8é = £76, and since 


B#1 then o= 0. 


Hence from (3) | d= aBd=a’'d. 

Since d+0, v=, 

and hence =p eee} 

Since p is odd B may be replaced by B’, so that B transforms both 7, and 7; 
into themselves. Hence there are two H, invariant in the whole G,:,., con- 


trary to hypothesis. 


15. gH, and q=—1(modp). Here the H,, must be non-cyclic. Neither 
T, nor T, can be transformed into any of its powers, except the first, by 


ib. or.C. 
The set of relations : 
Poe de Dea diy Bo) TB sel, 5 (ttl) Creed. 


eee LT Meu IN PD eT ues OE One To. 


32 M. O. TRIPP: GROUPS OF ORDER p’¢’ 


do not furnish a type, for 
GPU BMD Oral and BIATLG Ae CA Dente 


whence it follows that 7, = 7). 

If A transforms as above, neither B nor C can transform 7; and T, in a dif- 
ferent way from that above {cf. § 11 (4)}. Hence if a type exists in our sup- 
posed case A must be permutable with 7, and 7,, and then so far as the matter 
of isomorphism is concerned BC= CB, since A corresponds to the identical 


isomorphism. The only type then, that we can have, has the relations 
fof. § 11 (4)} 


AST A mee STs aah OT C2aT. 
ATTA IT MC RATUA in oT mn OG en 


TL 
G32 HAVING AN INVARIANT I, ; AND MORE THAN ONE HT. 


16. General considerations. If the elements of /7,, are all transformed by 
T, T,, or T,, we get the same elements in different order. Each element of 
an H7, corresponds to an isomorphism of //;. It follows, then, that q must 
be a divisor of the order of the group of isomorphisms of H,,. The orders of 
the groups of isomorphisms for the various types of 7, are given by WESTERN, 
l. c., pp. 211-216. 

If G,s,. contains p/Z,, then the 7, must contain an /Z,, each element of which 
is commutative with an H_,. 

If G,,,. contains p* 7, then the /7,, must contain an HZ, each element of which 
is commutative with an 7. 

If S represents one of the elements of 7; commutative with an 1, mentioned 


above then S1T7S = T* 
and hence repaid (be aes 1a Ws 
Since our H_, is invariant TST is an element of 17, and, therefore, 7*~ is also. 


Hence k = 1(mod q’) 


thus making S and 7 commutative. 
If the #7, are non-cyclic then we must consider two cases: 


(i) [Simin Eon ats bis pera MSs isl 
(ii) Seeipoientle [Sheri We Ped bec Ud 
In case (i) 


Sa(T ST) = TA Yand))) SA(T8T) = Teas 


M. 0. TRIPP: GROUPS OF ORDER p’¢? 33 


Now just as in the above cyclic case 
a=b=1 (modq). 
Hence 7,, 7, and S are all commutative. In case (ii) 
eel lon imc le Lay; and oped Ahi When lead Hed Meee 


Therefore T, T* belongs to H, which is impossible. Hence case (ii) is excluded. 
In considering each type of H, we will divide into cases according to the 
number of da . contained in G92 : 


17. HH cyclic, say A”*=1. The order of the group of isomorphisms is 
p’(p —1), and hence 
p=1 (modq). 


pIT,s. The f, 2, each of whose elements is permutable with the elements of 
an H,, must be {A?}. Hence 


A? TA? = th 
and since { A } is invariant in the G,,,. 
T7AAT = A’. 


This leads to two cases, according as a belongs to the exponent q or q’ mod p’. 


Now Tear) = Ae AP 
Hence a=1-4 kp’. 
Therefore a?=(1 + kp’)? =1 + kgp’(mod p*)=1 (mod p’), 
or at = (1 + kp’)? =1 + kq’p*(mod p*) = 1 (mod p’). 


In either case k = 0 (mod p) and hence A and 7’ are permutable, contrary to 
hypothesis. We evidently obtain the same result if the #7. are non-cyclic. 
pH. Proceeding in the same way as above we find that no type of Gs. 
exists in our supposed case. 
pH. For cyclic H,, we have 


TIAAT= A’. 
We get two types of G,,,. according as a belongs to the exponent qg or q° 
(mod p*). 
For non-cyclic 7, we have the relations 
TAS IAN hig oe Doe A T= AP, 
as in preceding work we may assume one of the exponents, say 6 = 1, while 


a belongs to the exponent g mod p*. Hence we get one type of G,,,» which 
is the direct product of an 7, and an £/,. 


34 M. 0. TRIPP: GROUPS OF ORDER p*q’ 


18. H,=[A” = B?=1,AB=BA]. The order of the group of iso- 
morphisms is p*(p—1)*. Hence p=1 (mod q). 

pi. The IT, with whose elements those of an Hf, » are commutative 
is either { A’, B} or {A}, where {A} is typical of the pH, {AB*} 
(«=0,1,2,---,p—1). If we take the former case, then 


(1) A>~TA*=T, BOTB=T. 


Since there are p cyclic H/,, in 7, one at least is permutable with 7. Suppose 
this is {A}. Then 
(2) THAT == A* 


and, as in the preceding, there are two cases. rom (1) and (2) we see that 
ap =p (modp’). Hence 
a=1+ kp. 

This makes G,,. Abelian, for 

at = (1+ kp)! = 1+ kgp(mod p’) = 1 (mod p’), 
or 

a? =(1+kp)” =1+kq’p(mod p?)=1 (mod p’). 
Hence in either case k = 0 (modp). Likewise it may be shown that no type 
of G,2 exists in the case of non-cyclic H,,. 

Next let us take 7 permutable with the elements of {A}. One at least, of 
the pH, {A”B},(4=0, 1, 2, ---,p—1) is permutable with 7. Taking 
this as { 6} we have 

DT Be aD 


We thus get two types of G,,,. according as a belongs to the exponent q or q* 
(mod p). Each is the direct product of {A} and {B, 7}. Concerning the 
latter group, see HOLDER, Mathematische Annalen, vol. 43, pp. 357-9). 
Let us take the non-cyclic /Z,, in which 7, 7 are permutable with A. 
We may now assume the relations : 


hgh aH ASAT od Moped oe ON Beat 
Hence ADT bec a ME RT ball 9 re Sate HU og LE eo Ke va 
Therefore akp = kp (mod p’) 
and hence k=0 or a=l. 
Whence it follows that in either case our relations take the form : 
LD dt De Ape ag Md 8 de 


Here again we may assume one of the exponents, say b=1 (modp). We 
thus get one type of G,;,. which is the direct product of {A}, { T,} and 
{#, 7}. 


a ee ee 





M. 0. TRIPP: GROUPS OF ORDER p*q’ 35 


19. p?H.. The H, whose elements are permutable with the elements of an 
H,, is either { A?} or {B}. The case of A” being permutable with the ele- 
ments of an 7, may be shown impossible just as in the preceding section. 


If we take TOBT=B 


then, since there are p cyclic H,., { AB*}, one, at least, is permutable with 
T. We may take this as { A}, and then 


gf BT: has 


thus giving us two types of G,.. Each is the direct product of {6} and 
anor |. ‘ 
For the Hf, » non-cyclic we have the relations (cf. § 18): 


d Beda Bs be a bey ee a a 


We may consider = 1 (mod p’), thus giving us a single type of G,,,., the 


direct product of { 7,}, {A, 7,} and { B}. ey 


m0: p ET s. First take the fal , cyclic. One, at least, of the p cyclic H,. is 
commutative with 7’ and this may be takenas {A}. Of the pH one, at least, 
is permutable with 7' and this may be taken as {B}. Hence we have the 
relations 


TAAT=A%, T-BT= B. 


a may belong to the exponent g or exponent g’ mod p’. 6 may belong to the 
exponent g or exponent q’ mod p’. Accordingly we have four cases to consider. 


a a primitive root of af‘=1 (mod p’), 


1 
@) b a primitive root of 6} = 1 (mod p). 


a may be thought of as any one of the primitive roots of 

a’=1 (mod p’), 
and then there are g — 1 types of G,,,. corresponding to the g — 1 values of 5. 
(2) a a primitive root of a’ =1 (mod p’), 


b a primitive root of b°=1 (mod p). 


On transforming with 7* [a taking g(q— 1) values] we get 6” in place of b 
and a’ [y taking g — 1 values] in place of a. Hence considering b as any one 


primitive root of 
b’=1 (modp), 


36 M. O. TRIPP: GROUPS OF ORDER p’q’ 


we get g — 1 types corresponding to the g — 1 values of a. 


a a primitive root of a” =: 1 (mod p’), 


(3) : 


b a primitive root of b}= 1 (mod p). 
This gives gy — 1 types of G,,,. corresponding to the g—1 values of b. 


@ a primitive root of a” = 1 (mod p’), 


(4) 


b a primitive root of 6° =1 (modp). 


This gives g(q — 1) types corresponding to the g(qg — 1) values of b. 
Non-cyclic H,,. Proceeding as in the cyclic case we may assume the relations 


THAT — At, THB BY OTAAT AB 
Transforming, 

em fhe Az BY fb ae —_ Avett bspy BB vka+bmy 
and 

Te fia A* BY (i the = Avztapry B bka+bmy : 


Hence we have the congruences 


(1) psb = asp (mod p’), (2) ak=bk (mod p). 
If a+6 (mod p) 
then s=0 and k=0 (mod p). 


If a=b (mod p) 


then 7, transforms every /7/,, in {A*} into itself, and every H, in { A”B} 
into itself. Hence our relations take the form ; 


TAT —A*, © UBT = BY) OP VAT en A= 7 ee 


6 and £ cannot both be = 1 (mod p), for then there would not be p*#7,. Like- 
wise we cannot have both a anda=1(modp). If 6, 8 = 1 we can change 
generators so as to assume b = 1(modp). Our relations may now be written 
in the form 


THAT A At, | TARTAR. er AT An on! Deere 


If a = 1 (mod p’) and hence a + 1 we get g — 1 types of G,,,., each being the 
direct product of {7\} and {7,, A, B}. Ifa=1,a+1 we have only one 
type. Ifa, a=+1 then we can keep 7; fixed and take 7 7% in place of T,, so 
that A is transformed into itself provided r is chosen properly. 


M. 0. TRIPP: GROUPS OF ORDER p*q’ 37 


2h |(AP= fh — CP =) AB BA,AC=]CA,BC=CB]. The 
group of isomorphisms is of order p?(p—1)?(p+1)(p?+p+1). 

pH,,. We must have p=1(modq). The 7, with whose elements the 
elements of an HH, are permutable may be taken as il, dosh 

For cyclic H, Dear = Ayand 1 Bla BT iss permutable with 
(p +1), viz. fA } and { AB*}. Since 


p?>=1 (mod g) 


one, at least, of the p’ remaining JZ, independent of A and B, must be permut- 
able with 7. Taking this as {C} we have 


Airtel Ci IS tae 


This gives two types, each of which is the direct product of H,={A, B} 
and 1 ,.=—{C, T}. The H, are treated by HOLDER, Matematische 
Annalen, Vol. 43, pp. 357-9. 

Non-cyclic H,,. Here T,, T, are commutative with A, B. We may assume 
the relations 


Oli GO", al =a3G Beas. 
Whence nba teases Boh Avy 
and ie nee tee poe AS, 


Hence we have aa=a and a8=f (modp). If a+1 then 2=0 and 
8=0. Ifa=1 then 7, transforms every 7, into itself so that in either case 
we can write our relations : 


TOT Oo ie OTS OT: mn C2 


and we can assume one of the exponents, say a = 1, thus giving us one type of 
G 2 Which is the direct product of { 7,} and {A, B, C, T,}. 


22. p’H,,. First we consider p = 1 (mod q). 

Cyclic H,,. The H, with whose elements T is permutable may be taken 
as { A}. 

If q¢> 2 then among the p’+ p other H, there are at least two permutable 
with 7’, and these two may be taken as (B } and {C}. Hence we have the 


relations : 
TIAT=A, T“BTl= B, TS OT == G?. 


We must have a, b + 1 for if either were congruent to 1 (mod p) there would 
not be p?H.. We must consider three cases. (1) a and 6 both primitive roots 
of a2=1(modp). This gives 3(q°— ¢ + 2) types of G,.2 [ef. § 4 (ii) (8)]. 
Each is the direct product of {A} and {B, C, 7}. (2) a and 6 both primitive 


38 M. O. TRIPP: GROUPS OF ORDER p’q’ 


roots of a’ =1(mod p). This gives 4(q + 1) types [§ 4 (ii) ()], and they 
are direct products as in (1). (3) Let a be a primitive root of av = 1 (mod p) 
and 6 a primitive root of at=1(modp). In this case we have g — 1 types 
corresponding to the g—1 primitive roots of a? =1 (mod p), and these are 
direct products as above. 

Non-cyclic H}:. (¢>2). TY, and T, are commutative with A. A similar 
consideration to that in § 20 shows that we may assume the relations 


(1) pi ead weer. Dy Dl be Dk OT aa Ce TT; BT cates 
a and a cannot both be congruent to 1 (mod p), and likewise for 6 and 8. We 
may assume one exponent, say b, congruent to 1. Then if 
a=1, a,8=1, 
we have }(q + 1) types, each the direct product of {A}, {7} and {C, B, 7)}. 
If a, «+1 we may keep 7, fixed and change the second generator 7) so that 
T, transforms C into itself. We, thus, get one type in which 6,2=1; a, B 
belong to the exponent g (mod p). 
q=2and ZH, cyclic. Besides { A} either none or at least two HZ, are per- 
mutable with 7’. If the latter is the case, then corresponding to (1) above we 
have two types; one in which 
T32CT= C%, TOBT = B, 
and a second type in which 
et Pe CO, LAB Tee 
Corresponding to (2) we have the single type with 
hath Os bE Bs Dot Bil ee bs 
and to (3) also a single type with 
POT =e or bere Nhat EMM Toft) 


If no other H, besides { A} is permutable with 7’, then it is easily seen that 
we must have 


TABT = C. 
Hence TUBE =T'CT= ABC’, 
also T3BT = Asta Bor (a+b 
and B=THABT = Artabt+eatad? Parad? (/2ab+b9 


If T? and B are commutative, we have 


T-(BC)T= BOC, 





M. O. TRIPP: GROUPS OF ORDER p*q’ 39 


showing that the H, {BC}, is permutable with 7 contrary to hypothesis. 
Hence the only possibility is that 7 is the lowest power of 7' permutable with 
B;; and, therefore, we have the congruences 


(1) a(1+b+a+4+06?)=0 (mod p); 
(2) a(a+b*)=1 (mod p); 
(3) b(2a +b?) =0 (mod p). 


Ifb = 0 then { 7, A, B} is an ZH, 


op: With an invariant IT. and having one 
H, viz., { A}, invariant. Hence there is a second H, say { B} invariant in 
H,,.. Therefore we can assume «= 0 when 6=0. Our supposed G,,,2 is 


the direct product of {A} and {B, C, 7}. Hence we have exactly the same 
case as in § 4 (ii), that is, we have one type of G‘,:,. with 


a=b=0, =—1, p=4m43. 


If 6 = 0 then (2a + 6’) = 0 from (3), and substituting the value of 6? in (2) we 
find that 


and hence p is of the form 4m +1. Therefore 


p+pt+1=3 +4+4k. 


If the 44/77, are transformed in k cycles of 47, each, that is, in each cycle the 
4/7, are transformed cyclically by 7’; then, since 3/7, cannot remain fixed, 
there is a cycle consisting of 2//,, i. e., some H, say { B} first goes into {C’} 
and then {C’} goes back into {#} under transformation by 7’. In every 
case, then, there is a cycle of 2/7, when all the (p* + p + 1) are transformed 
by T. This means that we may assume a=0 andb=0. This shows a con- 
tradiction, and hence there is no type for b + 0. 

q=2and H, non-cyclic. There can be no case in which only one #, viz. 
{ A} is permutable with 7, or 7), for if 


Ty Bi C,; 
then TOT = B, 


so that { BC} is permutable with 7. Our relations, accordingly, must be of 
the form (a). One type is given by the relations 


PeaO ee CO OT ART a. Bt OT OT = O35) PRT BR. 


This G,, is the direct product of { 7,}, {A} and {7, C, B}. A second 
Gs, in which {7,} is not invariant has the relations 


PCM CWT BT RY eh CTs One BIS = B-, 


40 M. O. TRIPP: GROUPS OF ORDER p’*q? 


23. p*H, and p= —1modgq. Here we take g> 2, for when q = 2 the 
congruences p = + 1 (mod q) are identical. 
Cyclic H, Se De ff, whose elements are commutative with 7’ may be taken 
as {A}. No other /, can be permutable with 7’, for the congruences 
a” =1 (mod p) and a? == 1) (mod py 


cannot have primitive roots since here p= 1 (mod q). Since there are 
(p?+p+1)H, in A, there is at least one H,, permutable with 7’, and evi- 
dently this H, may be taken as {B, C}; it cannot be taken as {A, B}. 
Evidently we have the following relations : 
TAUAT =A {US Teele, SERA OM R= Ferd C. 
Just as in § 5 we see that a= —1, b=#+4i%. There are two types accord- 
ingly as ¢ (the Galoisian imaginary) is a primitive root of 
i” =1 (mod p) or oft” =1 (mod p). 
In the latter case we must have 
p=-—1 (mod q’). 
These groups are the direct products of { A} and { B, C, T}. 


Non-cyclic H,,. A is commutative with 7, and 7,. We may have one type 
of G,,. with the relations : 


LB GC, Te Ol e= Beas [b=i?+i], 
ips Jol) Ae ee A be OS EA Ol 
This is the direct product of {7,}, {A} and {71, B, C}. There can be no 
other type (§ 7). 


24. pH, and p= 1(mod q). Let us first take g > 3 and H,, cyclic. Then 
at least 3H, are permutable with 77, since there are p?+p+1H,. Hence we 
must have the relations : 

TIAT = A*, TT" BT = B", PECT eG 


We may now have the following cases : 


(1) a, 6, call primitive roots of 6° =1 (mod p) 
(2) DO Cis 5S Sey nor b= AL sc ee 
(3) aand b « “ “6 SF =] “6 
c CC TOEAL Ote ceil De 
(4) a a co 66 G2? == “6 


b, eC “ ‘“ a4 62 =] CCAS 


M. O. TRIPP: GROUPS OF ORDER p°q’ 4] 


For case (1) we may set 
b=a*,c=a” (modp) (#,y=+k;1=0,1,2,---,qg—1;k=1,2,---p—1): 
Hence our relations above become 
TAAT = A, TT BT = Be, IB OU MEO 
To determine the number of types represented by these relations we set 
WO et Bae (2=lq+k). 
We may now get two distinct equivalences by taking z so that 
(i) w=1, (ii) zy =1 (mod gq’). 

First, ze =1 mod ¢ and 7,= 7’, A, = B, B, = A, C=C. Therefore 
Ty'A,T, = Aj, Ty BT, = By, Ty CT, = Cs". 

Second. vy =1(mod gq’) and 7,=7", 4,=C, B,= B, C= A. There- 
eee At, TO BT = ber CT, = Ce. 

Each pair of values (2, y) furnishes corresponding pairs (z, yz) and (va, v). 
Each of these three pairs gives the same type of group. The number of dif- 


ferent types is equal to the number of non-corresponding pairs. 
Now replace the numbers x, y, z, v by their indices (mod gq’), that is, we let 


n=g", y=9", 2g", v=g-" (mod g’). 


where g is a primitive root of q’?; and 2, y, take the values 
0, 1, 2, 3---[¢(g—1)—1]. We can now replace our triad of corre- 
sponding pairs by (2, ¥)),(— 2%» Yo — %)9 (%—Y%» —Y%) Letl=—y, 
m=x,,n=y,—x, [mod (q?—q)]. The triad of corresponding pairs now 
becomes 


(m, —/), (—m,n), (—n, l) 
and we have the congruence | 
1+m-+n=0 [mod (q’—q)]. 


The number of types depends on the number of solutions of this congruence. 
Let « be the number of triads (/, m, n) satisfying the congruence (disre- 
garding order) in which all three constituents of the triad are different, 8 the 
similar number in which two only are equal, and y the similar number in 
which all three are equal. IEf 
g = 1 (mod 3) 


then qg —q=0 (mod 3) 


and hence =o, 


42 M. 0. TRIPP: GROUPS OF ORDER p’q’ 


for the solutions are 


l=m=n=0, eae: eC Ace, [ mod (q* — q)]. 


3 
If q=2 (mod 3) 
then g—q#0 (mod 3) 
and therefore y=1. 


If two only of the numbers /, m, are equal, the congruence may be written 
L+2m=0 [mod(q’?—q)]. 


m may have any value except 0, 4(¢’?—q), [2(q’— ) ], and for each value 
of m there will be one value of /. Therefore if 


q= (mod 3) 
then B=7—q-3, 
and if q=2 (mod 3) 
then B= ¢—q-—1. 


The total number of solutions of all kinds of the congruence, considering the 
order of the constituents in each triad, is (q’—q)? Hence we must have 


6a+38+y=(9—g). 
If ¢ = 1 (mod 3) then substituting in the above 
a= $(¢'— 29 — 29° + 3g + 8), 
and if g = 1 (mod 38) we find that 
a= $(¢q' + 29° — 29° + 3q + 2). 


Let a, = the number of solutions a when one constituent of the triad is con- 
gruent to 0, and a, the number of solutions 2 when this is not the case. Hence 
a, is the number of solutions of 


1+ m=0 [mod (q’*—q)}], l+=0, m0, 
and excluding / =m =4(q’?—q). Therefore 








If q=1 (mod 3) 


M. O. TRIPP: GROUPS OF ORDER p’q’ 45 


we have a =a—a,=}(9q*— 29 — 5¢° + 674 12), 
and if q=2 (mod 3) 
then a%=a—a,=—4(¢*— 294° — 59° + 674+ 8). 


The a, solutions give a, types of groups; while the a, solutions give 2a, 
types of groups, since each triad in this case furnishes two distinct sets of cor- 
responding pairs. 

The triads (/, m, m) and (—/, —m, — m) give the same set,of correspond- 
ing pairs. If m =4(q°— q) the triads just named form the same solution ; 
but the other triads go in pairs, each pair furnishing one type. Hence from 
the 8 solutions mentioned above we get 


From y we get, when g = 1 (mod 3), two types; one corresponding to the 
triad (0, 0, 0) and asecond corresponding to the triad [4(q’— ¢7),4(¢— 7), 
-4(q@—4)]: while the triad {4[2(q*— 9), $[2(¢—9)], $[2(@—9)]} 
gives the same type as the second named above. If g = 2 (mod 3) we geta 
single type corresponding to the triad (0, 0, 0). 

In summary, then, for g = 1 (mod 3) the number of types is 


ee 
$ (9° — 29° — 5g’ + 6g + 12) +2—7 +2 = 4(9¢'— 2¢° + 4q°— 389+ 6), 


and for g = 2 (mod 3) the number of types is 


ares 
TO4 p= (gt 29° +4q°—39+ 2). 





ag 29 — oq +69 4-8) +9 —q—2+ 


Case (2) in which a, 6, c are all primitive roots of 6’ = 1 (mod p) may be 
treated in a way similar to the above and the number of types obtained will be 
exactly the same as is obtained by WESTERN (I. ¢., p. 287) for Gs,, viz. : 


aban Hale, 
6 





if q=1 (mod3), 


and ted if qg=2 (mod 3). 


Case (3) in which a and 6 belong to the exponent q’ and ¢ to the exponent 
q(mod p) leads to the relations 


TA ede TR Teepe ie Tt COT aCe. 


44 M. 0. TRIPP: GROUPS OF ORDER p*q’ 


The first two of these relations give $(q°>—q-+ 2) types of G'». (§ 4) and 
since c may have g—1 values, the whole number of types of Go is 
(q—1)[3(¢—¢ +2)]. 

Case (4), where a belongs to the exponent gq’? and 5, ¢ to the exponent 
q( mod p), gives, in a similar way, g(¢g —1)[3(q+1)] types. 

Non-cyclic H,(q>3). We may assume the relations 


eT Alas Mer Ved ye Wee PACT aie, 
a 
i IES ER 2 ere, LD ba DENA Cices 1,CT, = CY Ase 
We show now that our relations may be so changed that we can assume J, p, 
v, l, m,n=0(modp). Transformation of A, B, C by TT, and 7,7; 
shows the relations : 


a=b or A#=0, a=c or l= 
a=b or »=0, e=b or m= (mod p). 
a=c or v=0, c=b or n=0] 


If a, b, ¢ are all congruent (mod p) then every H is transformed into itself 
by T, and since 7), transforms 37, into itself, our relations (a’) may be made 
to take the required form. If a, 6, ¢ are not all congruent then two of them, 
say a and 6, are incongruent and hence A=0, »=0. We must now consider 
two cases (i) @ =, (il) a@ +c. 
In (i) c $ 6 and hence m =0,n=0. This gives for the transformation of 
A and C the relations 
UAL ee tAS: bres BUM TOK 
TNA AsG, 1G lie Caae 
Since a = c and g > 2 we may write our relations 
fhe RS Ea 2 be Uhre Wey M3 3/246 TENG T Sate 
TI AT, = As, T; BT, = B*A*, SA EM CY”. 
Transformation easily shows that we may assume » = O and hence relations 


(a’) take the required form. 
In case (ii) / = 0, » = 0. Here we must consider two subcases : 


(1) c#b, (2) omb. 


For (1) we see that n =0, m=0 and hence our relations (a’) take the required 
form. The subcase (2) may be easily shown to reduce to the required form 


M. O. TRIPP: GROUPS OF ORDER p*q* 45 


just as in (ii). Hence in every case we may write our relations (a’) 
DS AT ny A®, TB Tce\ BY, PCT as C 
TAT, = As, TT BT, = es T;1 CT, =C’. 


As in the preceding non-cyclic cases we can always make one of the expon- 
ents a, 8,y=1(mod p). Ifall three are congruent to 1 we get 





2 
4 
friar" types of G2, ifg =1 (mod 3); 
2 
and d z Z types, if g=2 (mod 3). 


Each of these types is the direct product of {7} and {7,, A, B, C}. 
If a= B=1, y #1 we get the same number of types as above. 
If only one of the exponents a, 8, y belongs to the exponent g (mod p) then 


there are 
ik - 4 
(SS ) (HE *) types forg=1 (mod 3) 


in 2 
and Gale +1) types for g=2 (mod 3). 


We must now consider what happens if g = 2 or 3 and there are not three 
independent 7, permutable with 7. If there are three such H, we may pro- 
ceed just as above. 

g = 2; we may assume that only one of the p? + p + 1H, is permutable with 
T. Suppose thisis {A}. Therefore 7“ AT= A’, T1BT=C. T? cannot 
be commutative with B, for then the group { BC} would be transformed into 
itself, contrary to hypothesis. For the same reason we cannot get a type of 
G,,. With our H,. non-cyclic. 

Since 7* is the lowest power of 7’ commutative with 6, any element of 
HT,; independent of A, we may write 


TBT? = TCT = A*B'C*. 
Therefore Liars T3 — T-2 CT? a At By Cut 
and B = Peay BUY Bi pas Aver axepay+ ae Py? +y2? Cyt? . 





Hence we must have the following set of congruences : 
e(a?+az+y+2)=0 
y(y+2)=1+ (mod p). 
z(2y+ 2)=0 


46 M. 0. TRIPP: GROUPS OF ORDER p'q’” 


If z=0 then {7”, A, B} is an H,,, with the H,= { A, B} invariant, and 
having 1/, = {A} invariant in this H,,. Hence there is a second 1, say 
{B}, invariant in this Z,,.. Therefore 


tae oy AAA Posh 
Accordingly we may assume x = 0 if z = 0 (mod p) and then y =— 1 (mod p). 
We thus get two types of G,, according as a belongs to the exponent 2 or 
exponent 4 (mod p) [ef. § 4 (ii)]. If 
z= 0 then Qy+27=0. 

Hence y°=—1 (mod p) 
and accordingly p is of the form 4m +1. Therefore 

pt+pt+iz=s3 (mod 4). 


Now the p’ + p/Z,, aside from {A}, must be transformed cyclically in sets of 
4H, each, for if there were 2/7, in any set we should have the case considered 
above in which z= 0. Accordingly we can not get a type of G,, with 2 = 0. 

q=3. Since it is supposed that there are not 3H, permutable with 7 there 
are none. Hence 


TIAT = B and TBI == PAL Borde 


If T* BT = A*B’ then {A, B, T} isan H,. having {A, B} as an invariant 
H,, and this H,, has p+ 1 H, which are permuted by 7, no one of them 
being invariant. 


Hence p=-—1 (modq). 


This contradicts the hypothesis that » =-+ 1 (mod q), and accordingly we must 
assume 7 1BT'=C’. If we suppose 7* commutative with A then we have 


TATA Bo) TABT SC, )\)) PACTS 


and the group { ABC} is invariant under 7’, contrary to hypothesis. 
Evidently the non-cyclic H,, leads to the same result as above. 
Let us suppose 7” is the lowest power of 7’ commutative with A. The 
p> +p +14, may be transformed in cycles of either 3H, or 9H. If there are 
3H, in any one cycle then with 


TCAT=B and TUBT=C 
we must have 7!CT'= A’, whence 
T 3 AT* = A’, LP BP eB TA OT = Ge, 
that is, each of the p* + p+ 14, is transformed into itself by 7°. Since a 


M. 0. TRIPP: GROUPS OF ORDER p*q’ 47 


cannot be unity it must belong to the exponent 3 (mod p). This furnishes one 
type of G.,,.. 
If every cycle contains 9H, then we have 


(a). p+p+1=0 (mod 9) 
Let p=%9k+a; k=0,1,2::-; Cet 2r ranis i mao 
Hence pPt+pt+1l=@’+a+1 (mod9). 


It is easily seen that v+ta+tils+0 (mod 9) 
and hence congruence («) is impossible, so that we cannot get a type in this case. 
25. pH, and p?’+p+1=0 (modq). 


Cyclic H,. Evidently q+ 2, and if g=3 the congruences p=1 and 
p’+p+1= 0 (mod g) are identical. Hence we need only consider the case 
in which g> 3. 

None of the 7, can be permutable with 7, for p = 1 (mod q). The p’+p 
+ 1£, must fall into sets of g or g’°H, each, the /Z, in each set being permuted 
cyclically by 7. We may assume 


Pera Des) Soeanda BL a C. 
for if aS Eien AS B® 
then the Z,..= {7, A, B} isa group already treated whose existence depends 


on the congruence p = — 1 (mod q), which is not true here. Hence we must 
have the relations : 


PAP eR Nr ATs Cl). TACT AtRe Cr: 


We must now consider two cases : 

(i) T¢ is commutative with C. 

(ii) 7° is the lowest power of 7 commutative with C. 

In case (i) we may proceed just as WESTERN does (loc. cit., pp. 240-4) thus 
getting a single type of G,.,. in which y, 8, « satisfy the relations : 


ys ALA +, 
ieee AS A PtP , 
as Arteti= 1, 


X, ”, ”* being Galoisian imaginaries of the third order and primitive roots of 


the congruence 
t= 1 (modp). 


48 M. O. TRIPP: GROUPS OF ORDER p*q’ 
Case (ii) also furnishes a single type, the relations being the same as in (i) 
except that ’, X”, p”’ are primitive roots of 
Av =1 (modp), 
thus requiring that p+p+1=0 (mod q’). 
The procedure in cases (i) and (ii) are so nearly alike that it is unnecessary to 
go through case (11). 
Non-cyclic H,,. We may have a single type of G,;,. with the relations : 
eA det: UUaew op by 6 TOL = 4 bee 
Lice Oe WIM esha Te) =e {og h Ou is 9 6: 
This group is the direct product of { 7,} and {7,, A, B, C}; a, B, y are 
determined just as in (i) for the cyclic case. 


26. ET eh GA Vee NS ee eer 


The order of the group of isomorphisms of our /Z, is 24 and since q is a divisor 
of this order we must have g=3. Evidently there must be 4H7,. T is com- 
mutative with A? since { A? is a characteristic subgroup. 

Our #, contains three cyclic H,, viz.: 


{A}, {AB} = {1, AB, A’, A°B}, {B} = (1, B, A’, A*B}. 


Cyclic H,. T must (1) either be commutative with each of the above , or 
(2) permute them cyclically. We cannot have the first case, for 


TAT = A’ 
leads to one of the two congruences 
ve=1 or a®°=1 (mod 4). 


and the only value a can have in each case is unity. Hence //, is invariant in 
an #7,, and, therefore, in the G,, contrary to hypothesis. 
In the second case we may take 


TAAT- By) TORT AA) wore ae 


Hence TAT = TAABT or Sa ea aged 
= BAB or BAB 
=A. 


From this we that 7* is always commutative with A and B. We thus get a 
single type of G,,. 


M. O. TRIPP: GROUPS OF ORDER p*q’ 49 


Non-cyclic H,,. There is a single type with the relations : 
ee Abe ——sbs' 1 BI = AB, 
TVAT oui A, TAB = B. 
This G,, is the direct product of {7} and H,,={7,, A, B}. 
It may be noted that we cannot have a G,. with 
i eet Ae De a a ie Ao 
for here the order of the group of isomorphisms of H, is 8 and, therefore, is not 
divisible by q. 
27. ef eee Ale Dea ie A == APs plodd), 


Pp 


The order of the group of isomorphisms of H,, is p?(p — 1); and since this 
must be divisible by g, we have 
p=1 (mod q). 
(i) pi... 
Cyclic H,,. The H,, with whose elements 7’ is permutable is either {A} or 
{A?, B}. In the former case we have 


TIAT=A. 


{A?} is an H, permutable with 7. There are p other #,, viz.: {AB}, 


(k=0,1,2---p—1), forming a conjugate set. On account of the congruence 


p=1 (mod q) 
one of these H,, say { 8}, is permutable with 7. Hence 
| TA BT = Be. 
Therefore B2iAT= AUBoOT= ATR; 
also Alien 8) Avent) br Aaa LAP Boe AMT Bo, 
Hence a=1 (mod p). 


This makes A, B commutative with 7’, contrary to hypothesis. The same 
result follows for non-cyclic H,,. In the latter case we have 


Ti AeT= A, T-3BT=B. 


There are p cyclic H,, { AB*}, (k= 0, 1, 2, ---, » —1) and, as in the preced- 
ing, one of these, say { A}, is commutative with 7’ and hence 


ie AcL => A*; 


where « is a primitive root of 


at or a” =1 (mod p’). 


50 M. O. TRIPP: GROUPS OF ORDER p’q’ 


Theretore LAAT = Ae, 

But T- APT = A?, 

whence a=1 (mod p) 

or a=1+hikp. 

Therefore (l+kp)* or (1+%kp)”=1 (mod p’) 


which requires that k = 0 (mod p), and accordingly A, B, and T are all com- 
mutative, contrary to hypothesis. Non-cyclic H_,, leads to the same result. 

(ii) pH». 

Cyclic H,. The H, with whose elements 7’ is commutative may be taken 
as { B} since it cannot be { A?}. One of the p cyclic H,,, say { A}, is per- 
mutable with 7. Hence 

TAAT= A 
This furnishes two types of G',:,. according as a belongs to the exponent q or 
exponent q? (mod p’). 
Non-cyclic HT,. We may always assume the relations 


ina ean Tp Ale 
for if DA Ly ae 


then on transforming A with 7,7, = 7,7, we see that either a=1ork=0. 
Hence we obtain a single type of G’,,,2 which is the direct product of {7,, A, B} 
and {7',}, since we may assume by a proper change of generators that a = 1 
(mod p’). 

(ii) pH. 

Cyclic H,. One of the p cyclic H,., say {A}, is commutative with 7’ and 
one of the p#Z,, say { B}, is also commutative with 7. Hence 


TjAAT=A*, TOBT=B?. 


Therefore T- B24 BT = TOAPH Te Avety . 
also TABOAABT= B?T3ATR = BA*B’ = Avert), 
Hence 6 =1 (mod p) which cannot be true with p*/7. 


Non-cyclic H,. Just as in the cyclic H, we may write 
f bear Uf Ae Ba Ud eae 
and upon transforming A and B with 7, we may write 
W bespie CIB Cre We Ly Bi yea Be 
Hence Tt TAL, 1p, — A fpaxax—1) Fpkax 


M. 0, TRIPP: GROUPS OF ORDER pq? 51 


and fhe as katy | LD a Aut —spexe—]) Be x 
Therefore k=0 or a=1 
so that we may assume Ah AL Sate 


In like manner we may show that 
oid vesnra as 


and just as in the cyclic case above 8 =1; hence we fail to get a type of 
G2 With p°H7_.. 


Some | Aves Peal, AB= BA, AC=CA, C“*BC=AB]. 

(7) pH. We must have p = 1 (mod q). 

Cyclic H,. As the H,, whose elements are commutative with 7’ we may 
take {A, B}. Tis permutable with the p+1H,, {B} and {AB}, 
(k=0, 1, 2,---p—1). Since there are p?+p-+1H, and since p?= 
(mod gq), one of the remaining p’H,, say {C’}, must be commutative with 7; 
and so we may write 


Bin OF LEH Be 
From BT = TB we get 


(Oso j}(C EC yen £O)( CIBC) 
But AB=C"BC and CATC=TCO- 4 
Hence AB and TC-**! are commutative. Therefore 
ABTO #1 = TC-HAB= ATA BC = A* BIC-*+', 
Hence | a=1 (modp), 


which makes C and 7 commutative, an impossible condition under our 
hypothesis. 

From the above it is evident that there can be no type of G,;,. with H,, non- 
eyclic. 

(ii) p’? H, and p= 1 (modq). 

Cyclic H, x. che a , whose elements are commutative with 7 may be (1) the 
invariant characteristic subgroup {A } or, (2) some other /7,, say { B}. 


In case (1) To AT aa As 


If g + 2 among the p’ + p remaining /Z,, there are at least 2/7, commutative 


with 7. Hence 
PU BT mtb. NOT = GY, 


From C"*BC=AB, 


52 M. 0. TRIPP: GROUPS OF ORDER p*q’ 
we get, since CT=TC and TAO Sh On ten, 

TC BOT = T3 ABT = AB* = CT BTC” = C” BC? = A” Be, 
Hence ab=1 (modp). 


Accordingly a and 6 are related and 6 = a*—' or at". We, therefore, get two 
types of G,,,.; according as a belongs to the exponents ¢ or exponent q” (mod p). 

Non-cyclic H,. As in the case of cyclic H, we can assume for one of the 
non-identical elements of H7,., say 7,, the following relations 


TABT =B, TCT. =C™, 


where a belongs to the exponent g(modp). The most general transforma- 
tions of B and C’by T,, are 


T> BT,= A: B80,  1T;1OT, = A B+’, 


Hence f bal Roos oa Maga Ee argh ee POA E RL 
and Tele BUT ete Deed oars 
Since a1 we have y=0, a=0. 


Similarly by transforming C’ we find X=m=0. Hence we can assume rela- 
tions as follows: 


TB. = Bt, TACT = C0", TART = Be) Cre 


If «+= 1 (mod p); then, by taking a proper combination of 7, and 7, in place 
of T,, B can be transformed into itself. Hence we may assume « = 1 (mod p). 
We, therefore, get one type of G,,., the direct product of {7,} and 
{1, 4, B, OC}. 

If ¢g = 2 and two of the p’+ pH, besides {A}, are commutative with 7’, the 
above procedure is applicable. Hence we need consider only the case where 
q = 2 and none of the p? + pH, are commutative with 7. 

The p’? + pf,. besides {A}, are transformed by 7 in cycles of 2 or 4H, each. 
If any one cycle has 2, in it then we may assume the relations : 


TAAT — Al.) TB? ei Giec i 


Hence {heed 88 batea ta PRS 8 
Therefore b=+1 (modp). 
If b=+1 then diene of Biome = Crap Oe 


which is contrary to hypothesis. We thus see also that non-cyclic H, is im- 
possible with g=2. Hence b= —1 is the only permissible value. This 


M. 0. TRIPP: GROUPS OF ORDER p’q? 53 


furnishes one type of group with the relations : 
je ONES Tbe ee ee Cr, as Gee De 


Next let us suppose that no cycle contains 2H7,. Consequently every cycle of 
the p’ + pH, contains 47, and we may write our relations as follows : 


TAT =A,  TIOBT=0,  T7OT= ABC’. 


Therefore TB? = T3CT= ABC 
and T3BT? = TOT? = A% BY’ C*+*, 
also B=T BT =T CT? = At Bete (Bete 


where a,, a, are functions of a, b,c. Hence we must have the congruences : 
2be+c=0 (modp), b?+ be? =1 (modp). 


If c =0, then 6? = +1, a case already considered. Hence we need consider 
onlyc=+ 0. Therefore 


26+c=0 (modp), b> =—1 (modp), 
so that » must be of the form 4n+1. Therefore 
p+tpt+1=3 (mod4). 


This means that one cycle must contain 2/7,, contrary to hypothesis. 
We now consider case (2) in which 


Lee ts Demonia 


If ¢ + 2, then there are 2/7 , besides {8}, permutable with 7. Since {A} is a 
characteristic HZ, it must be one of our 2/7,. The other one we may call {C’}. 


Hence (by Palen ees PAT tA {hal Ou aga ky 
From C'BC = AB we get on transforming with 7 
US OCulB GT Ar B 
and since C7'= TC” we have 
Gee ee Ge = Carb Cara RAB 
Hence c=a (mod p). 


We thus get two types of G,., according as a belongs to the exponent q or 
exponent qg’ (mod p). 
Non-cyclic H,. It is easily seen that we may assume the relations 


Tig Dla. A ae ie Gh =O 
Us tease Ty Ad =n As OT == AS Be Ce 


54 M. 0. TRIPP: GROUPS OF ORDER p’¢’ 


By a proper change of generators we can assume a = 1; and accordingly 
«= y = 0 (mod p) so that we get a single type of G,..., the direct product of 
{7} and {7,, A, B, C}. Just as in the cyclic case we have 


b=2z (modp). 


If g=2, then 1H,, {B}, being permutable with 7 there are p’?+ pH, — 
remaining. But {A} is also permutable since it is a characteristic H,. Taking 


out {A} and {B} we have left 
p+p—1=1 (mod 2). 


Hence among the p? + p — 1 H, one at least, say { C’}, is permutable with 7, 
and hence the case g = 2 offers nothing new. 

(iii) p>, and p = —1 mod q. 

Here we take g > 2 for if g = 2 the congruences 


p=1 (mod gq) and p=-—1 (mod gq). 


are identical. 

Cyclic H,,. Since {A} is a characteristic subgroup and p = — 1 (mod q), 
A and T must be commutative. No other H, can be commutative with 7. 
Hence we have the relations : 


TIAT=A, TOBT=C, T-CT=A*BCy, 


Proceeding just as WESTERN does (Il. c., pp. 251-3), we may show that we 
get two types of G,,,., according as (1) 7% is commutative with C, or (2) T” 
is the lowest power of 7 commutative with C. In both cases, we have 


a=0, B=—1, y=r+r (mod p); 


where A is a Galoisian imaginary and a primitive root, in case (1) of x? = 1 
(mod p), and in case (2) of *= 1(modp). In the latter case we must have 
p = —1(mod q’). 

Non-cyclic H,,. We have one type of G‘,:,. with the relations 


ead eee TD Pens Da Ol p= Bah Gy 
T, AT, =A, Ti bT, = B; AAT ES BOE TT ES 
y having the same value as in case (1) above. This is the direct product of 
{T,} and { 71, A, B, C}. 
(iv) p’H,». By Sylow’s Theorem we have p*=1 (mod q). Since the 


group of isomorphisms of this /7,, is of order p*(p —1)?(p +1) and since g 
must divide this order we must have p = 1 (mod q@). 


M. 0. TRIPP: GROUPS OF ORDER p’¢’ 55 


Cyclic H,. Ifq> 2 then at least two of the 7/,, besides {A}, must be per- 
mutable with 7’, so that we have 


TAT = A‘, TIBI = B’, MAM OM LEM Bi 
Let us transform C-!'BC = AB with T. Therefore 
TAC3ABCT=TO3ABT 
and hence A™ B = A*B?, 
so that be=a (mod p). 


If a, 6, or c belongs to the exponent g’ (mod p), then the other two do, so 


that we may put 
OG =) and ae 


Therefore e+y=t1 (mod q’). 


Neither B nor C can be put in place of A, for A and its powers are the 
only invariant elements of 7,;. and C can be interchanged. Accordingly 
the number of types is the number of solutions of x + y = 1 (mod q’) subject 
to the condition that v, y + 0, 1. 

If g =2 the congruence has no solutions satisfying our conditions. If gq> 2 
there is one solution 
q+ 

2 





L=y= 
for which x = y, and $(q° — 2g — 1) for which xy. Thus we get 


Wied deny god tL 
Fee ne cn 
types. 
If a belongs to the exponent q (mod p) then 6 and ¢ do, and the number of 


types is the number of solutions of 


a+y=1 (modq), 
and is, therefore, equalto *—~—+1=4——. 


Non-cyelic HT, 2» We may assume the relations 
beds WY Age, Ge (Me ee ME Tara Poe Oy MER OE 
Pe ea Ae dip I 311 ap Toy Crees (as 
By a proper change of generators we may make 
a=l and then By =1 (modp). 


Also just as in the cyclic case a = be (mod p). 


56 M. O. TRIPP: GROUPS OF ORDER p’*q’ 


If 8 and ¥ belong to the exponent g (mod p) we get $(q — 1) types of G2... 
If B=y=1 we get $(¢q—1) types the direct product of { 7,} and 
HL Awol 510 es 
= 2. Since the case of 3/7, commutative with a non-identical element of 
H, , is impossible, we need consider only the case in which one H, is commuta- 
tive with Tor 7,, TZ). 
Cyclic lah 2. Since {A} is commutative with 7’ we have 


TAT= A*, TABT=HC. 


T must transform the p?+ p&,, aside from {A}, in cycles of 2H, or 4H, each. 
If any one cycle contains 2H, then we have 


TAB xe TACT =i"; 
We must consider two cases: 


(1) b=+41, (2) b=—1 (modp). 


Now TAO BOT =x Pat A Bale 
In case (1) therefore, A =e ee 
Hence =—1 (modp) 

and TAX? Y BCT = AMPUBC, 


which is contrary to hypothesis. 
For case (2) in which 6 = — 1 we find from 


Daa Givoet AD 
that AC = AC 


Hence a = 1 which is impossible with p’ 7». 
If no cycle contains 27, then just as in § 28 (ii) we fail to get a type. 
Non-cyclic H,, is evidently impossible. 


Ve 
Gy... HAVING NEITHER AN INVARIANT HH, NOR AN INVARIANT Z,,. 
The only possible orders for groups of this kind are 72 and 108 (§2). 


29. G,. The 4H, have in common an H,, which is invariant in the G 
thus leading to a factor group T',,. This T,, has 3H, or 14. 

(1) If there are 3H, in our T’,,, these H, have in common an H, which is 
invariant in the I',,, corresponding to which we get in G,, an invariant H,,. 

(2) If there is 1H, in our [,,, then there is an invariant /Z,, in our G,,; and 
this H,, has 1H, or 3H,. If the H,, has 1/, it is invariant in the G,,, contrary 


72? 


M. 0. TRIPP: GROUPS OF ORDER p’q’ 57 


to hypothesis. If the #,, has 3H,, then there are only 3H, in our G,,; and these 
3H, have in common an /, invariant in the G,,. An invariart H/, and an 
invariant ZZ, lead to an invariant H,, of the G,,. 

Corresponding to the invariant /,, obtained in the two cases above, we get a 
factor group I’, having 1H; and hence G,, has an invariant H,,. This H,, 
must have 4H,, for if it had only 1H, this H, would be invariant in the G,,. 
The 4/7, of the invariant H,, have an H, (invariant) in common. Hence we 
get a factor group I’,, with 44, and, therefore, 147, which is non-cyclic. Hence 
Hf,, has an invariant H7,,. This H,, has 1H, or 3H; and these are all the /,’s 
there are in H,,. If there are 3, in /,, and, therefore in H,,, these H, have 
an H, in common and invariant in ,,. We thus get a factor group I’,, having 
1H, and, therefore, our invariant H,, has an invariant #7, containing 1/4). 
Therefore, 7, is invariant in H,, and accordingly in G,,, contrary to hypothesis. 
It follows then that our invariant H,, has only 1H, which is also invariant in the 
Hi, and G,,. 

The invariant /, cannot be cyclic, for then we should have an H, invariant 
in the H,,, which was excluded above. In our supposed case, then, we must 
have an invariant H, but not an invariant H,. The 3H, or 9, have in com- 
mon the invariant H,. The H, cannot be of the type 


i Bestt Web Ad BAB AS, 


for the invariant £7, is cyclical. 
9H,. The largest subgroup J in which any one of these 9, is invariant is 
of order 8. If the 9H, are Abelian, then the invariant 7, has an H, invariant 


in the G,,. Therefore, there is no type of G,, with Abelian , in our supposed 
case. 


In our putative case the 97, must be of the type 4* = 5? =1, B1AB= A’; 
and we may take as our invariant non-cyclic /, 


(1, A?, B, A?B}. 


Cyclic H,. T must transform A’, B, AB cyclically, while 7 is permutable 
with each of them, for {7%} is invariant in G‘,,. Hence we may assume 


T7“A*?T=B and TOBT = A’B. 
From A?7'= TB we have 


(1) ATA? = TBA? = TA*B. 
Since { 7’, A’, B} is our invariant £,,, 
(2) AWTA 17 AB: 


Values of x, y, z must be so chosen that (1) and (2) harmonize. 


58 M. 0. TRIPP: GROUPS OF ORDER p’q* 


T* is not permutable with A; for if it were 7* would be permutable with 
each element of an H/,, and hence { 7°, 7,} = an H,, with only 1H,, which is 
impossible. Since {7°} is invariant, A-*7°A = T°. 

With reference to (1) and (2) it is evident that we need to consider only the 


following values : 
Pin—he WA! Robt PesvOee Ls: te Oa 7 he 


y =2z=0 being excluded. This gives six cases to be tested. 


(i) ripen b yeah” Ts Ay 
Hence (2) gives ATTA ei TA?, 
and 7: te WY Need 
not agreeing with (1). 
(ii) Cents y= 0, z=1, 
AT LA weil is. 
Na a As EASA be Ue 
again contradictory. 
(iii) Bex di; youl, 2x1, 
ATA = TAtB, 
ATA es TA 


again contradictory. © 
(iv) c= 8, yl, z2=0, 


A“TA = T*A?, 


AS Aten ae 
which agrees with (1). 


(v) eto Vi, zg=1, 
AA TA met? Bs 
ATA? = L, 


which contradicts (1). 

(vi) La 8, yee els, gee 1; 
AT? TA et ASB 
A“ TA* = TAB’, 

which agrees with (1). 

Cases (iv) and (vi) furnish the same type of group; for if in (vi) we replace 
T by T? we get the same relation as in (iv). We thus get a single type of 
G,, defined by the relations 

A‘= B= T° =1, BI AB= A’, 
AT =) T“ BT = A’*B, A GTA fee el A 


M. 0. TRIPP: GROUPS OF ORDER p'q° 59 


Non-cyclic H,. As above our H, are of the type 
At stay Dh) ee 1s BAB = A’, 
and our invariant HH, may be taken as | 
ool phe at a Ue Ass BD ad oe 


_ Now 7,, 7, cannot both be permutable with A’ or B, for then G,, would have 
an invariant H,, which is not allowable. Hence we may assume 


TI A?T = B, T BT = AB. 
If Dales B or as EY; 
then on transforming A? by 7, 73 (x = 2 in Ist case, 1 in 2nd case) in place of 
T,, we find that A? and, therefore, also B and A’ B are permutable with 7, 7%. 


Hence the elements A’, B, A? B may always be taken permutable with 7'. 
Since { 7} is our invariant H, we must have 


Arce d  Agese ke or hes 


The first case is impossible with 9/7, for our G,, would be the direct product 
of {7} and {7,, A, B} and accordingly could not contain more than 3Z,. 


Therefore Age Art ls 
Since {7\, 7,, A’, B} is our invariant H,, 

(3) ee Abe A esd 
also 

(4) FiWmeid i Pe Rana dbs Eat a 


Values of a, x, y, 2 must be so taken as to make (3) and (4) agree. Trans- 
formation of (3) by A, when x = 1, leads to a contradiction ; and if « = 2 we 
find thata=0,y=2=1. Hence 


| Arde Ace wh Avda 
Now { 7,, A, B} is an H,, with 3H, (cf. Burnside, Theory of Groups p. 104). 
Since Leads at A 


we see that there are H, not included in the HZ, above; and hence there must be 
9H, in our G,,,. 
We, therefore, have a G‘,, with the defining relations : 


tee Bat BTA Bis At wy Pet 42 TB, 
TA BT Aes. At TAS 7? APB, 
EOS a Migs i Ae RBI eal 


60 M. 0. TRIPP: GROUPS OF ORDER p’q* 


3H,. I, the largest subgroup in which an H, is invariant, is of order 24. If 
our HH, are Abelian no two elements of our invariant H, can be conjugate in 
I= H,,; for in this ,, every element of our invariant H, is permutable with 
every element of an H,. Hence each element of our invariant 7, is invariant 
in G,,, giving us an invariant 7, in G,,, which is not allowable. 


The only type of #, for us to consider is 
At‘= B=1, BIAB= A’. 
As in the case of 9H, our invariant H, may be taken as {1, A’, B, A? B}. 


Cyclic H,. T* must be permutable with each element of the above invariant 
Since there must not be an invariant H7, in G,, we must have 


TIAVT= 5B and T“BT = A*B. 


fal 


4: 


{T, A’, B} is our invariant H,,. Hence 


(5) ATA = T*A™” B 
also 

(6) ATA? = TA’?B 
and 

(7) Apel Aes 


Testing the six possible sets of values x, y, 2 just as in the cyclic H, with 9H, 
we see that no set satisfies the relations (5), (6), (7) and hence no type of G,, 
exists in our supposed case. 

Non-cyclic H,. From the discussion of the non-cyclic H, with 9H, it is 
evident that only a single type of G,, is possible under our conditions. The 
defining relations are 


BofA BH1, BIAB= 4, Tan Bo 7, ee 
AT A—-T?AB, Th=TT, BATRA Te 


This G,, is the direct product of {7\} and {7,, 4, B}. Since the latter is an 
H,, having 3H, and 4#7,, our G,, must have 3H, and 4H,. 

We have found above 2H,, having 4H,. This agrees with MILLER, 
Quarterly Journal of Pure and Applied Mathematics, vol. 28, p. 283. 


30. Gi... The 4#,, have in common an #, invariant in the G‘,,,. From 
the invariant H, we get the factor group I’,, with 4H, and, therefore, 14, ; 
corresponding to which in G',,, we have an invariant H/,,. Hence there are 
3f, or 9H,. 

If there are 3H, they have in common an H, which is invariant in the G,,,. 
Corresponding to this invariant H,, we get a factor group I',, with 1#,. 
Therefore G,,, has an invariant /7,, with 1//,,; and hence this 1, is invariant 


108 


M. 0. TRIPP: GROUPS OF ORDER p*q’ 61 


in the G,,,, contrary to hypothesis. Accordingly the only case we need to 
consider is 9/7,. 

If our H,, are Abelian each element of the invariant H, must be invariant in 
the G,,,. Hence there could be only 14, or 3H,, both of which cases are 
excluded under our conditions. It follows, then, that in our supposed G’,,, the 
H,, must be non-Abelian. 

Cyclic H,. The factor group I’,, with 4H, mentioned above is isomorphic 
with the tetrahedral group; therefore, I',, has an invariant non-cyclic H,. 
From the isomorphism of I’,, and G,,, we see that the 9H, cannot be cyclic. 

Non-cyclic H,. Suppose the H,, are of the type 


AS = BP a 1} B3AAB= A’. 
Our invariant HZ, may be taken as {A} or {A*, B}. 

If it is the former then 7,, 7, cannot both be commutative with A, for then 
we would not have 9H,. Hence one of them, say 7,, must transform A thus: 
Ae ae, 

If we also have A begiva Wis peek >, Ros 


then by keeping 7; fixed and replacing 7, by 7,7, we see that 7,7, transforms 
A into itself. Hence we may assume 


Ti AT, =A. 


This, however, makes 7, common to the 9H, and, therefore, invariant in the 
Gog) a case already excluded. 

Suppose our invariant H, = {A*, B}. 2H, of the invariant H, are per- 
mutable with one of the elements 7,, 7. Therefore we may write 


POR AST a A, MRAM BS ire da 
We may also write 


TA AT, = AMB, Tz BT, = A*B*. 
Now Tp TAABL T, = Avat%e Booras 
and TT; A°BT, T, = Ametse Boe +a , 


If a+ 8 then c = 0 and 6 = 0 (mod 8); so that 2H, are invariant under both 
T, and T,. If «=8 then all the H, in our invariant H, are invariant under 
T,; and since 2H, are invariant under 7), we have again 2H, invariant under 
both 7, and 7,. Hence we may write our relations as follows : 

i Ae ea eer iste. 

LAS) aon ae AP WSLS dee 


62 M. 0. TRIPP: GROUPS OF ORDER p’q’ 


If aand a are both congruent to 2 (mod 3) then, keeping 7; fixed and replacing 
T, by T,T,, we see that 
(7, 1) A(T, L,) = 4. 

Hence we may assume a2=z,a=1 unless a=a=1. The latter case is 
excluded, for then we can make 6 or 8 = 1 so that we have an #H, invariant in 
the G,,,. Hence we may always assume a= 2,a=1. 

If b =1, then { 7,} is an invariant H, of G,,,, which has been shown im- 
possible in our supposed case. Henceb=2. If now 8 = 2 then keeping 7, 
fixed and replacing 7, by 7, T, we see that 


(2,7, BUTT) = B. 
Hence we can assume 8 = 1 and so we have the relations 
TIA Tent AP, dhe Sess hel LACS atlas Tl; BY, aie 
Since { 7,, 7,, A*®, B} is an invariant H,, we have 
a Nove Mo's Brea bod bt badd etd 


Let us transform 7, by AB= BA‘. 
(i) Let b= 1. Hence 


Abo Atel) (Ag aes 
also Bo At AD = tT e Age ee 
Therefore x + 2 =a and y+2=y(mod3). These congruences being con- 
tradictory, it is impossible to get a type in our supposed case with b= 1. 


(ii) 6=0. Hence 
ACD eT BAS ep As) ee ee 


and A LAB ee ee 
Therefore « = « + 2 (mod 3), again contradictory. It follows, then, that we 


cannot get a G',, with the H,, taken as above, in our supposed case. 
Let us take the H,, of the type 


SuB=O=1, ABH BA, AC= CAs) C1 2G eee 


Without loss of generality we may take our invariant H,as {A, B}. Then 
our invariant H,, is {A, B, 7,, T,}. Proceeding just as in the other non- 
Abelian case it is evident that relations for our invariant H,, may be taken as 
follows : 


TAT) = AMC TRB TNR Tt Ae me 


M. 0. TRIPP: GROUPS OF ORDER p*q’” 63 
Since { 7,, Z,, A, B} is our invariant Z,, 
Conk Oe te At Be, 
Let us transform using the fact that B C= CAB. 


(i) b=a=1. 
Hence Colheal BO. bd A.D 
and Daa Gal OAL ea le At oe. 


whence 2 = 0 (mod 3), an impossible result. 


(ii) a=0, b=1 ’ 

Hence Cm Baal) Dee 1 ALB, 
and Barta CuwliG Abie dA, Bots 
again impossible. 

(iii) = Ls5 a 0. 

Hence : C1 BIT BC = T. A* B’, 
and Dawaw Cm, OAD ef Alte Be 


again contradictory. 
Evidently we cannot hvea=b=0. 
There is, then, no Gy. having neither an invariant H, nor an invariant H,,. 


Vi 


sg HAVING AN INVARIANT /7,, AND ALSO AN INVARIANT HZ. 


G 


P 


32. Since the subgroups #,, and H,, have no element in common except 1, 
we may apply Theorem IX, p. 44, BuRNsIDE’s Theory of Groups, viz.: 

If every operation of G transforms H into itself and every operation of H trans- 
forms G into itself, and if G and H have no common operation except identity; 
then every operation of G is permutable with every operation of H. 

If p = 2, g having any value, there are ten types of G‘,,,. arising from taking 
the direct product of the five types of /7,, and the two types of H,,. 

Likewise there are ten types of G,.,. when p is odd. 


VITA. 


Myron Owen Tripp was born near Grand Haven, Mich., March 19, 1874; 
entered Sodus Academy, N. Y., in 1891, graduating therefrom in 1893; grad- 
uated from the Northern Indiana Normal, 1895; Teacher in Sodus Academy, 
1895-6; Principal of the Sodus Point, N. Y., public school, 1896-7; Student 
in Engineering, Michigan Agricultural College, 1897-8; Principal of the 
Sodus Public School, 1898-9; Special Student in Mathematics and Education, 
Michigan University, 1899-1900; A. B. Indiana University, 1901; Instructor 
in Mathematics, Physics, and Chemistry in the National Farm School, Doyles- 
town, Pa., 1901-2; Graduate Student in Mathematics and Education, Cornell 
University, 1902-3; Teacher of Mathematics in the College of the City of 
New York, 1903 until the present time, and at the same time Graduate 
Student in Columbia University; Member of the American Mathematical So- 
ciety. The writer takes this opportunity of expressing his thanks to Professor 
Cole for suggestions and helpful criticism. 


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